Proof of fundamental lemma of calculus of variation.












7












$begingroup$


Suppose $Omega$ is an open subset of $mathbb{R}^n$ and let $L^1_text{Loc}Omega$ denote all locally integrable functions on $Omega$ and $C^{infty}_0Omega$ for smooth functions whose support lie in $Omega$. My teacher tell me the following statement:



Suppose $fin L_{Loc}^1Omega$ and
$$int_Omega fvarphi=0,forallvarphiin C^infty_0Omega$$
Then $f=0text{ a.e.}$



It is known as fundamental lemma of calculus of variation. My teacher told me it suffices to prove this statement holds for the case $f$ is continuous. But I find it's not easy to deduce the lemma from the case $f$ is continuous. Could someone tell me how to do this or how to prove the lemma directly? Thanks a lot!










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Hint: One (possible) intermediate step is to prove the lemma if $varphi$ are characteristic functions. Then you can deduce the lemma just by approximating your characteristic functions. You can also find the answer in most textbooks and also on the internet. I dont think, that the requirement to $f$ being continuous is a significant simplification - you can leave out the a.e., I dont think that anything else interesting happens.
    $endgroup$
    – Daniel
    Nov 11 '14 at 23:45








  • 2




    $begingroup$
    Continuity of $f$ simplifies the proof greatly, because for continuity if $f$ is not the 0 function then there exists an interval $[a,b]$ where $f$ is strictly positive, and if $phi$ is a smooth function with support contained in $[a,b]$ (which can easily be built) then $int_{mathbb{R}}fphi=int_a^bfphi>0$, which contradicts the hypothesis. Now if we assume we can take characteristics as $phi$s, and $f$ is not 0 but is integrable, …
    $endgroup$
    – MickG
    Jul 11 '15 at 18:45






  • 2




    $begingroup$
    … then we certainly have a measurable set $A$ where $f>0$, and if $phi$ is $chi_A$ then $int fphi=int_Af>0$, which would contradict the hypothesis if we assume the integral is zero for $phi$ taken to be measurable characteristics. The next step would be to approximate such a characteristic with smooth functions, i.e. to prove that if $chi$ is a characteristic there exist $phi_n$ smooth compactly supported functions that converge to $chi$ pointwise a.e.. We would probably need to assume $chi$ is the characteristic of a bounded subset, but that doesn't harm the above argument …
    $endgroup$
    – MickG
    Jul 11 '15 at 18:49






  • 1




    $begingroup$
    … since if there is a set $A$ then we can find a bounded subset of $A$. Assuming we have such $phi_n$s, we apply dominated convergence, and here I guess we _do_ need boundedness on the part of $A$, because the integrable domination would be provided precisely by $fchi$, assuming the $phi_n$s are all less than one, which can be given by normalization, or anyway a multiple of $fchi$. So we are left with proving the existence of $phi_n$. So how do we go about proving that if $chi$ is the characteristic of a measurable set then we can find smooth compactly supported $phi_n$s which converge…
    $endgroup$
    – MickG
    Jul 11 '15 at 18:51






  • 1




    $begingroup$
    … to $chi$ pointwise a.e.? We have a domination for $|chi-phi_n|$, so $L^1$ convergence implies pointwise convergence a.e. by Dominated convergence (again), so we need to prove $mathcal{C}^infty_0$ is dense in $L^1$ (i.e. the set of complactly supported smooth functions is dense in $L^1$). As you can see, @Daniel, the proof is considerably longer without continuity. Is what I said above right or is there any detail (or major mistake) to fix?
    $endgroup$
    – MickG
    Jul 11 '15 at 18:54
















7












$begingroup$


Suppose $Omega$ is an open subset of $mathbb{R}^n$ and let $L^1_text{Loc}Omega$ denote all locally integrable functions on $Omega$ and $C^{infty}_0Omega$ for smooth functions whose support lie in $Omega$. My teacher tell me the following statement:



Suppose $fin L_{Loc}^1Omega$ and
$$int_Omega fvarphi=0,forallvarphiin C^infty_0Omega$$
Then $f=0text{ a.e.}$



It is known as fundamental lemma of calculus of variation. My teacher told me it suffices to prove this statement holds for the case $f$ is continuous. But I find it's not easy to deduce the lemma from the case $f$ is continuous. Could someone tell me how to do this or how to prove the lemma directly? Thanks a lot!










share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Hint: One (possible) intermediate step is to prove the lemma if $varphi$ are characteristic functions. Then you can deduce the lemma just by approximating your characteristic functions. You can also find the answer in most textbooks and also on the internet. I dont think, that the requirement to $f$ being continuous is a significant simplification - you can leave out the a.e., I dont think that anything else interesting happens.
    $endgroup$
    – Daniel
    Nov 11 '14 at 23:45








  • 2




    $begingroup$
    Continuity of $f$ simplifies the proof greatly, because for continuity if $f$ is not the 0 function then there exists an interval $[a,b]$ where $f$ is strictly positive, and if $phi$ is a smooth function with support contained in $[a,b]$ (which can easily be built) then $int_{mathbb{R}}fphi=int_a^bfphi>0$, which contradicts the hypothesis. Now if we assume we can take characteristics as $phi$s, and $f$ is not 0 but is integrable, …
    $endgroup$
    – MickG
    Jul 11 '15 at 18:45






  • 2




    $begingroup$
    … then we certainly have a measurable set $A$ where $f>0$, and if $phi$ is $chi_A$ then $int fphi=int_Af>0$, which would contradict the hypothesis if we assume the integral is zero for $phi$ taken to be measurable characteristics. The next step would be to approximate such a characteristic with smooth functions, i.e. to prove that if $chi$ is a characteristic there exist $phi_n$ smooth compactly supported functions that converge to $chi$ pointwise a.e.. We would probably need to assume $chi$ is the characteristic of a bounded subset, but that doesn't harm the above argument …
    $endgroup$
    – MickG
    Jul 11 '15 at 18:49






  • 1




    $begingroup$
    … since if there is a set $A$ then we can find a bounded subset of $A$. Assuming we have such $phi_n$s, we apply dominated convergence, and here I guess we _do_ need boundedness on the part of $A$, because the integrable domination would be provided precisely by $fchi$, assuming the $phi_n$s are all less than one, which can be given by normalization, or anyway a multiple of $fchi$. So we are left with proving the existence of $phi_n$. So how do we go about proving that if $chi$ is the characteristic of a measurable set then we can find smooth compactly supported $phi_n$s which converge…
    $endgroup$
    – MickG
    Jul 11 '15 at 18:51






  • 1




    $begingroup$
    … to $chi$ pointwise a.e.? We have a domination for $|chi-phi_n|$, so $L^1$ convergence implies pointwise convergence a.e. by Dominated convergence (again), so we need to prove $mathcal{C}^infty_0$ is dense in $L^1$ (i.e. the set of complactly supported smooth functions is dense in $L^1$). As you can see, @Daniel, the proof is considerably longer without continuity. Is what I said above right or is there any detail (or major mistake) to fix?
    $endgroup$
    – MickG
    Jul 11 '15 at 18:54














7












7








7


2



$begingroup$


Suppose $Omega$ is an open subset of $mathbb{R}^n$ and let $L^1_text{Loc}Omega$ denote all locally integrable functions on $Omega$ and $C^{infty}_0Omega$ for smooth functions whose support lie in $Omega$. My teacher tell me the following statement:



Suppose $fin L_{Loc}^1Omega$ and
$$int_Omega fvarphi=0,forallvarphiin C^infty_0Omega$$
Then $f=0text{ a.e.}$



It is known as fundamental lemma of calculus of variation. My teacher told me it suffices to prove this statement holds for the case $f$ is continuous. But I find it's not easy to deduce the lemma from the case $f$ is continuous. Could someone tell me how to do this or how to prove the lemma directly? Thanks a lot!










share|cite|improve this question











$endgroup$




Suppose $Omega$ is an open subset of $mathbb{R}^n$ and let $L^1_text{Loc}Omega$ denote all locally integrable functions on $Omega$ and $C^{infty}_0Omega$ for smooth functions whose support lie in $Omega$. My teacher tell me the following statement:



Suppose $fin L_{Loc}^1Omega$ and
$$int_Omega fvarphi=0,forallvarphiin C^infty_0Omega$$
Then $f=0text{ a.e.}$



It is known as fundamental lemma of calculus of variation. My teacher told me it suffices to prove this statement holds for the case $f$ is continuous. But I find it's not easy to deduce the lemma from the case $f$ is continuous. Could someone tell me how to do this or how to prove the lemma directly? Thanks a lot!







real-analysis calculus-of-variations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 11 '14 at 23:39









Michael Hardy

1




1










asked Nov 11 '14 at 23:26









RichardRichard

13218




13218








  • 2




    $begingroup$
    Hint: One (possible) intermediate step is to prove the lemma if $varphi$ are characteristic functions. Then you can deduce the lemma just by approximating your characteristic functions. You can also find the answer in most textbooks and also on the internet. I dont think, that the requirement to $f$ being continuous is a significant simplification - you can leave out the a.e., I dont think that anything else interesting happens.
    $endgroup$
    – Daniel
    Nov 11 '14 at 23:45








  • 2




    $begingroup$
    Continuity of $f$ simplifies the proof greatly, because for continuity if $f$ is not the 0 function then there exists an interval $[a,b]$ where $f$ is strictly positive, and if $phi$ is a smooth function with support contained in $[a,b]$ (which can easily be built) then $int_{mathbb{R}}fphi=int_a^bfphi>0$, which contradicts the hypothesis. Now if we assume we can take characteristics as $phi$s, and $f$ is not 0 but is integrable, …
    $endgroup$
    – MickG
    Jul 11 '15 at 18:45






  • 2




    $begingroup$
    … then we certainly have a measurable set $A$ where $f>0$, and if $phi$ is $chi_A$ then $int fphi=int_Af>0$, which would contradict the hypothesis if we assume the integral is zero for $phi$ taken to be measurable characteristics. The next step would be to approximate such a characteristic with smooth functions, i.e. to prove that if $chi$ is a characteristic there exist $phi_n$ smooth compactly supported functions that converge to $chi$ pointwise a.e.. We would probably need to assume $chi$ is the characteristic of a bounded subset, but that doesn't harm the above argument …
    $endgroup$
    – MickG
    Jul 11 '15 at 18:49






  • 1




    $begingroup$
    … since if there is a set $A$ then we can find a bounded subset of $A$. Assuming we have such $phi_n$s, we apply dominated convergence, and here I guess we _do_ need boundedness on the part of $A$, because the integrable domination would be provided precisely by $fchi$, assuming the $phi_n$s are all less than one, which can be given by normalization, or anyway a multiple of $fchi$. So we are left with proving the existence of $phi_n$. So how do we go about proving that if $chi$ is the characteristic of a measurable set then we can find smooth compactly supported $phi_n$s which converge…
    $endgroup$
    – MickG
    Jul 11 '15 at 18:51






  • 1




    $begingroup$
    … to $chi$ pointwise a.e.? We have a domination for $|chi-phi_n|$, so $L^1$ convergence implies pointwise convergence a.e. by Dominated convergence (again), so we need to prove $mathcal{C}^infty_0$ is dense in $L^1$ (i.e. the set of complactly supported smooth functions is dense in $L^1$). As you can see, @Daniel, the proof is considerably longer without continuity. Is what I said above right or is there any detail (or major mistake) to fix?
    $endgroup$
    – MickG
    Jul 11 '15 at 18:54














  • 2




    $begingroup$
    Hint: One (possible) intermediate step is to prove the lemma if $varphi$ are characteristic functions. Then you can deduce the lemma just by approximating your characteristic functions. You can also find the answer in most textbooks and also on the internet. I dont think, that the requirement to $f$ being continuous is a significant simplification - you can leave out the a.e., I dont think that anything else interesting happens.
    $endgroup$
    – Daniel
    Nov 11 '14 at 23:45








  • 2




    $begingroup$
    Continuity of $f$ simplifies the proof greatly, because for continuity if $f$ is not the 0 function then there exists an interval $[a,b]$ where $f$ is strictly positive, and if $phi$ is a smooth function with support contained in $[a,b]$ (which can easily be built) then $int_{mathbb{R}}fphi=int_a^bfphi>0$, which contradicts the hypothesis. Now if we assume we can take characteristics as $phi$s, and $f$ is not 0 but is integrable, …
    $endgroup$
    – MickG
    Jul 11 '15 at 18:45






  • 2




    $begingroup$
    … then we certainly have a measurable set $A$ where $f>0$, and if $phi$ is $chi_A$ then $int fphi=int_Af>0$, which would contradict the hypothesis if we assume the integral is zero for $phi$ taken to be measurable characteristics. The next step would be to approximate such a characteristic with smooth functions, i.e. to prove that if $chi$ is a characteristic there exist $phi_n$ smooth compactly supported functions that converge to $chi$ pointwise a.e.. We would probably need to assume $chi$ is the characteristic of a bounded subset, but that doesn't harm the above argument …
    $endgroup$
    – MickG
    Jul 11 '15 at 18:49






  • 1




    $begingroup$
    … since if there is a set $A$ then we can find a bounded subset of $A$. Assuming we have such $phi_n$s, we apply dominated convergence, and here I guess we _do_ need boundedness on the part of $A$, because the integrable domination would be provided precisely by $fchi$, assuming the $phi_n$s are all less than one, which can be given by normalization, or anyway a multiple of $fchi$. So we are left with proving the existence of $phi_n$. So how do we go about proving that if $chi$ is the characteristic of a measurable set then we can find smooth compactly supported $phi_n$s which converge…
    $endgroup$
    – MickG
    Jul 11 '15 at 18:51






  • 1




    $begingroup$
    … to $chi$ pointwise a.e.? We have a domination for $|chi-phi_n|$, so $L^1$ convergence implies pointwise convergence a.e. by Dominated convergence (again), so we need to prove $mathcal{C}^infty_0$ is dense in $L^1$ (i.e. the set of complactly supported smooth functions is dense in $L^1$). As you can see, @Daniel, the proof is considerably longer without continuity. Is what I said above right or is there any detail (or major mistake) to fix?
    $endgroup$
    – MickG
    Jul 11 '15 at 18:54








2




2




$begingroup$
Hint: One (possible) intermediate step is to prove the lemma if $varphi$ are characteristic functions. Then you can deduce the lemma just by approximating your characteristic functions. You can also find the answer in most textbooks and also on the internet. I dont think, that the requirement to $f$ being continuous is a significant simplification - you can leave out the a.e., I dont think that anything else interesting happens.
$endgroup$
– Daniel
Nov 11 '14 at 23:45






$begingroup$
Hint: One (possible) intermediate step is to prove the lemma if $varphi$ are characteristic functions. Then you can deduce the lemma just by approximating your characteristic functions. You can also find the answer in most textbooks and also on the internet. I dont think, that the requirement to $f$ being continuous is a significant simplification - you can leave out the a.e., I dont think that anything else interesting happens.
$endgroup$
– Daniel
Nov 11 '14 at 23:45






2




2




$begingroup$
Continuity of $f$ simplifies the proof greatly, because for continuity if $f$ is not the 0 function then there exists an interval $[a,b]$ where $f$ is strictly positive, and if $phi$ is a smooth function with support contained in $[a,b]$ (which can easily be built) then $int_{mathbb{R}}fphi=int_a^bfphi>0$, which contradicts the hypothesis. Now if we assume we can take characteristics as $phi$s, and $f$ is not 0 but is integrable, …
$endgroup$
– MickG
Jul 11 '15 at 18:45




$begingroup$
Continuity of $f$ simplifies the proof greatly, because for continuity if $f$ is not the 0 function then there exists an interval $[a,b]$ where $f$ is strictly positive, and if $phi$ is a smooth function with support contained in $[a,b]$ (which can easily be built) then $int_{mathbb{R}}fphi=int_a^bfphi>0$, which contradicts the hypothesis. Now if we assume we can take characteristics as $phi$s, and $f$ is not 0 but is integrable, …
$endgroup$
– MickG
Jul 11 '15 at 18:45




2




2




$begingroup$
… then we certainly have a measurable set $A$ where $f>0$, and if $phi$ is $chi_A$ then $int fphi=int_Af>0$, which would contradict the hypothesis if we assume the integral is zero for $phi$ taken to be measurable characteristics. The next step would be to approximate such a characteristic with smooth functions, i.e. to prove that if $chi$ is a characteristic there exist $phi_n$ smooth compactly supported functions that converge to $chi$ pointwise a.e.. We would probably need to assume $chi$ is the characteristic of a bounded subset, but that doesn't harm the above argument …
$endgroup$
– MickG
Jul 11 '15 at 18:49




$begingroup$
… then we certainly have a measurable set $A$ where $f>0$, and if $phi$ is $chi_A$ then $int fphi=int_Af>0$, which would contradict the hypothesis if we assume the integral is zero for $phi$ taken to be measurable characteristics. The next step would be to approximate such a characteristic with smooth functions, i.e. to prove that if $chi$ is a characteristic there exist $phi_n$ smooth compactly supported functions that converge to $chi$ pointwise a.e.. We would probably need to assume $chi$ is the characteristic of a bounded subset, but that doesn't harm the above argument …
$endgroup$
– MickG
Jul 11 '15 at 18:49




1




1




$begingroup$
… since if there is a set $A$ then we can find a bounded subset of $A$. Assuming we have such $phi_n$s, we apply dominated convergence, and here I guess we _do_ need boundedness on the part of $A$, because the integrable domination would be provided precisely by $fchi$, assuming the $phi_n$s are all less than one, which can be given by normalization, or anyway a multiple of $fchi$. So we are left with proving the existence of $phi_n$. So how do we go about proving that if $chi$ is the characteristic of a measurable set then we can find smooth compactly supported $phi_n$s which converge…
$endgroup$
– MickG
Jul 11 '15 at 18:51




$begingroup$
… since if there is a set $A$ then we can find a bounded subset of $A$. Assuming we have such $phi_n$s, we apply dominated convergence, and here I guess we _do_ need boundedness on the part of $A$, because the integrable domination would be provided precisely by $fchi$, assuming the $phi_n$s are all less than one, which can be given by normalization, or anyway a multiple of $fchi$. So we are left with proving the existence of $phi_n$. So how do we go about proving that if $chi$ is the characteristic of a measurable set then we can find smooth compactly supported $phi_n$s which converge…
$endgroup$
– MickG
Jul 11 '15 at 18:51




1




1




$begingroup$
… to $chi$ pointwise a.e.? We have a domination for $|chi-phi_n|$, so $L^1$ convergence implies pointwise convergence a.e. by Dominated convergence (again), so we need to prove $mathcal{C}^infty_0$ is dense in $L^1$ (i.e. the set of complactly supported smooth functions is dense in $L^1$). As you can see, @Daniel, the proof is considerably longer without continuity. Is what I said above right or is there any detail (or major mistake) to fix?
$endgroup$
– MickG
Jul 11 '15 at 18:54




$begingroup$
… to $chi$ pointwise a.e.? We have a domination for $|chi-phi_n|$, so $L^1$ convergence implies pointwise convergence a.e. by Dominated convergence (again), so we need to prove $mathcal{C}^infty_0$ is dense in $L^1$ (i.e. the set of complactly supported smooth functions is dense in $L^1$). As you can see, @Daniel, the proof is considerably longer without continuity. Is what I said above right or is there any detail (or major mistake) to fix?
$endgroup$
– MickG
Jul 11 '15 at 18:54










1 Answer
1






active

oldest

votes


















5












$begingroup$

I will now outline a proof I managed to come up with, so as to have this outline available online. It seems a web search does not present a proof that doesn't assume continuity on the part of $f$…



Continuous case



If $f$ is continuous, the steps are two:




  1. Prove the hypothesis stays true if you substitute smooth compactly supported (henceforth cs) functions with characteristics of balls; this is achieved via smooth transitions, which approximate characteristics of balls pointwise (but evidently not uniformly);

  2. $f$ is continuous, so if $f(x)neq0$ for some $x$, there exists a ball around $x$ where $f$ has constant sign; integrate $f$ times the characteristic of that ball and you should get something nonzero by what I just said, and zero by step 1: a contradiction; so $f=0$ everywhere.


General case



The proof here is much longer.





  1. Approximate continuous functions uniformly (and thus in $L^1$ and pointwise a.e.) by convolutions; in other words, define:



    $$rho_n(x)=left{begin{array}{cc}
    n^Ne^{frac{1}{1-|x|^2}} & |x|leq1 \
    0 & text{otherwise}
    end{array}right.,$$



    where $N$ is the dimension of the space $f$ is defined on (i.e., $f:Omegasubseteqmathbb{R}^Ntomathbb{R}$); then let:



    $$rho_nast u(x)=int_{mathrm{supp}(u)}rho_n(x-y)u(y)mathrm{d}y=int_{B(0,frac1n)}rho_n(y)u(x-y)mathrm{d}y.$$



    By the first expression, this is a smooth function; the support is the sum of the ball and the support of $u$, thus if $uin C_c(Omega)$, the support is compact; finally, using the second expression and a few manipulations, one can prove $rho_nast u-u$ has $L^1$ norm less than the supremum of $|tau_yu-u|_{L^1}$ over $|y|<frac1n$, where $tau_yu(x)=u(x-y)$; that $L^1$ norm reduces to integrating over the compact support of $u$, thus $u$ is uniformly continuous there, and this can be used to prove that sup tends to zero, and hence $rho_nast uto u$ in $L^1$; a subsequence thus converges a.e., everything here is bounded by the sup of $u$ which is continuous and compactly supported therefore bounded, so we have a domination and we conclude $frho_nast uto fu$ in $L^1$, so the integrals converge, but the LHS always integrates to 0;



  2. For any $frac1n$, one has a finite cover of a compact set $KsubseteqOmega$ by balls of radius $frac1n$, and thus a partition of unity, i.e. a collection of $phi_i^{(n)}$ which are in bijection with those balls, are each zero outside the matching ball, and sum to 1 on $K$ and less than 1 out of it; plus, they are continuous, so $fsumphi_j^{(n)}$ always integrates to 0; finally, those converge pointwise (at least a.e.) to the indicator of $K$, so since we have a domination we conclude the integrals of $fsumphi_i^{(n)}$ tend to the integral of $f$ over $K$ as $ntoinfty$, but those integrals are all 0 by step 1; hence $f$ integrates to 0 over all compact subsets of $Omega$;


  3. The Lebesgue measure is inner regular, so there is, for any measurable $MsubseteqOmega$, a sequence $K_n$ of compacts such that $mu(K_n)tomu(M)$; this easily implies $L^1$ convergence of the indicators of those compacts to that of $M$; thus, along a subsequence, there is pointwise a.e. convergence; so I should be able to deduce the integrals of $f$ over those compacts converge to that of $f$ over $M$, but I'm not so sure I can use dominate convergence here; perhaps I can just restrict myself to the sets below…;


  4. $f$ is measurable due to integrability, hence $F^+:={f>0},F_-:={f<0}$ are both measurable; their characteristics are approximated by characteristics of compact subsets as shown above; in those cases, I can apply Fatou if Dominated Convergence doesn't apply; Fatou surely yields $int_Omega f^+leq0$, as well as the integral of $f^-$; but those functions are positive, hence those integrals are 0, and they sum to $int_Omega|f|$, which is therefore 0; and bingo: this implies -- at long last!! -- $f=0$ a.e..



Can someone please verify these steps, and say if I've gone wrong at some point? In particular, in general case step 3, I'm afraid DC is unavailable in general since if $Omega$ is not bounded a constant is no domination, and besides I still have $f$ around which is only locally integrable, so I can use DC if $M$ is contained in a compact set, otherwise $f$ integrated over $M$ might well be infinite… so can I use DC to simplify things or do I have to use Fatou on the sets of point 4?






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    1 Answer
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    5












    $begingroup$

    I will now outline a proof I managed to come up with, so as to have this outline available online. It seems a web search does not present a proof that doesn't assume continuity on the part of $f$…



    Continuous case



    If $f$ is continuous, the steps are two:




    1. Prove the hypothesis stays true if you substitute smooth compactly supported (henceforth cs) functions with characteristics of balls; this is achieved via smooth transitions, which approximate characteristics of balls pointwise (but evidently not uniformly);

    2. $f$ is continuous, so if $f(x)neq0$ for some $x$, there exists a ball around $x$ where $f$ has constant sign; integrate $f$ times the characteristic of that ball and you should get something nonzero by what I just said, and zero by step 1: a contradiction; so $f=0$ everywhere.


    General case



    The proof here is much longer.





    1. Approximate continuous functions uniformly (and thus in $L^1$ and pointwise a.e.) by convolutions; in other words, define:



      $$rho_n(x)=left{begin{array}{cc}
      n^Ne^{frac{1}{1-|x|^2}} & |x|leq1 \
      0 & text{otherwise}
      end{array}right.,$$



      where $N$ is the dimension of the space $f$ is defined on (i.e., $f:Omegasubseteqmathbb{R}^Ntomathbb{R}$); then let:



      $$rho_nast u(x)=int_{mathrm{supp}(u)}rho_n(x-y)u(y)mathrm{d}y=int_{B(0,frac1n)}rho_n(y)u(x-y)mathrm{d}y.$$



      By the first expression, this is a smooth function; the support is the sum of the ball and the support of $u$, thus if $uin C_c(Omega)$, the support is compact; finally, using the second expression and a few manipulations, one can prove $rho_nast u-u$ has $L^1$ norm less than the supremum of $|tau_yu-u|_{L^1}$ over $|y|<frac1n$, where $tau_yu(x)=u(x-y)$; that $L^1$ norm reduces to integrating over the compact support of $u$, thus $u$ is uniformly continuous there, and this can be used to prove that sup tends to zero, and hence $rho_nast uto u$ in $L^1$; a subsequence thus converges a.e., everything here is bounded by the sup of $u$ which is continuous and compactly supported therefore bounded, so we have a domination and we conclude $frho_nast uto fu$ in $L^1$, so the integrals converge, but the LHS always integrates to 0;



    2. For any $frac1n$, one has a finite cover of a compact set $KsubseteqOmega$ by balls of radius $frac1n$, and thus a partition of unity, i.e. a collection of $phi_i^{(n)}$ which are in bijection with those balls, are each zero outside the matching ball, and sum to 1 on $K$ and less than 1 out of it; plus, they are continuous, so $fsumphi_j^{(n)}$ always integrates to 0; finally, those converge pointwise (at least a.e.) to the indicator of $K$, so since we have a domination we conclude the integrals of $fsumphi_i^{(n)}$ tend to the integral of $f$ over $K$ as $ntoinfty$, but those integrals are all 0 by step 1; hence $f$ integrates to 0 over all compact subsets of $Omega$;


    3. The Lebesgue measure is inner regular, so there is, for any measurable $MsubseteqOmega$, a sequence $K_n$ of compacts such that $mu(K_n)tomu(M)$; this easily implies $L^1$ convergence of the indicators of those compacts to that of $M$; thus, along a subsequence, there is pointwise a.e. convergence; so I should be able to deduce the integrals of $f$ over those compacts converge to that of $f$ over $M$, but I'm not so sure I can use dominate convergence here; perhaps I can just restrict myself to the sets below…;


    4. $f$ is measurable due to integrability, hence $F^+:={f>0},F_-:={f<0}$ are both measurable; their characteristics are approximated by characteristics of compact subsets as shown above; in those cases, I can apply Fatou if Dominated Convergence doesn't apply; Fatou surely yields $int_Omega f^+leq0$, as well as the integral of $f^-$; but those functions are positive, hence those integrals are 0, and they sum to $int_Omega|f|$, which is therefore 0; and bingo: this implies -- at long last!! -- $f=0$ a.e..



    Can someone please verify these steps, and say if I've gone wrong at some point? In particular, in general case step 3, I'm afraid DC is unavailable in general since if $Omega$ is not bounded a constant is no domination, and besides I still have $f$ around which is only locally integrable, so I can use DC if $M$ is contained in a compact set, otherwise $f$ integrated over $M$ might well be infinite… so can I use DC to simplify things or do I have to use Fatou on the sets of point 4?






    share|cite|improve this answer









    $endgroup$


















      5












      $begingroup$

      I will now outline a proof I managed to come up with, so as to have this outline available online. It seems a web search does not present a proof that doesn't assume continuity on the part of $f$…



      Continuous case



      If $f$ is continuous, the steps are two:




      1. Prove the hypothesis stays true if you substitute smooth compactly supported (henceforth cs) functions with characteristics of balls; this is achieved via smooth transitions, which approximate characteristics of balls pointwise (but evidently not uniformly);

      2. $f$ is continuous, so if $f(x)neq0$ for some $x$, there exists a ball around $x$ where $f$ has constant sign; integrate $f$ times the characteristic of that ball and you should get something nonzero by what I just said, and zero by step 1: a contradiction; so $f=0$ everywhere.


      General case



      The proof here is much longer.





      1. Approximate continuous functions uniformly (and thus in $L^1$ and pointwise a.e.) by convolutions; in other words, define:



        $$rho_n(x)=left{begin{array}{cc}
        n^Ne^{frac{1}{1-|x|^2}} & |x|leq1 \
        0 & text{otherwise}
        end{array}right.,$$



        where $N$ is the dimension of the space $f$ is defined on (i.e., $f:Omegasubseteqmathbb{R}^Ntomathbb{R}$); then let:



        $$rho_nast u(x)=int_{mathrm{supp}(u)}rho_n(x-y)u(y)mathrm{d}y=int_{B(0,frac1n)}rho_n(y)u(x-y)mathrm{d}y.$$



        By the first expression, this is a smooth function; the support is the sum of the ball and the support of $u$, thus if $uin C_c(Omega)$, the support is compact; finally, using the second expression and a few manipulations, one can prove $rho_nast u-u$ has $L^1$ norm less than the supremum of $|tau_yu-u|_{L^1}$ over $|y|<frac1n$, where $tau_yu(x)=u(x-y)$; that $L^1$ norm reduces to integrating over the compact support of $u$, thus $u$ is uniformly continuous there, and this can be used to prove that sup tends to zero, and hence $rho_nast uto u$ in $L^1$; a subsequence thus converges a.e., everything here is bounded by the sup of $u$ which is continuous and compactly supported therefore bounded, so we have a domination and we conclude $frho_nast uto fu$ in $L^1$, so the integrals converge, but the LHS always integrates to 0;



      2. For any $frac1n$, one has a finite cover of a compact set $KsubseteqOmega$ by balls of radius $frac1n$, and thus a partition of unity, i.e. a collection of $phi_i^{(n)}$ which are in bijection with those balls, are each zero outside the matching ball, and sum to 1 on $K$ and less than 1 out of it; plus, they are continuous, so $fsumphi_j^{(n)}$ always integrates to 0; finally, those converge pointwise (at least a.e.) to the indicator of $K$, so since we have a domination we conclude the integrals of $fsumphi_i^{(n)}$ tend to the integral of $f$ over $K$ as $ntoinfty$, but those integrals are all 0 by step 1; hence $f$ integrates to 0 over all compact subsets of $Omega$;


      3. The Lebesgue measure is inner regular, so there is, for any measurable $MsubseteqOmega$, a sequence $K_n$ of compacts such that $mu(K_n)tomu(M)$; this easily implies $L^1$ convergence of the indicators of those compacts to that of $M$; thus, along a subsequence, there is pointwise a.e. convergence; so I should be able to deduce the integrals of $f$ over those compacts converge to that of $f$ over $M$, but I'm not so sure I can use dominate convergence here; perhaps I can just restrict myself to the sets below…;


      4. $f$ is measurable due to integrability, hence $F^+:={f>0},F_-:={f<0}$ are both measurable; their characteristics are approximated by characteristics of compact subsets as shown above; in those cases, I can apply Fatou if Dominated Convergence doesn't apply; Fatou surely yields $int_Omega f^+leq0$, as well as the integral of $f^-$; but those functions are positive, hence those integrals are 0, and they sum to $int_Omega|f|$, which is therefore 0; and bingo: this implies -- at long last!! -- $f=0$ a.e..



      Can someone please verify these steps, and say if I've gone wrong at some point? In particular, in general case step 3, I'm afraid DC is unavailable in general since if $Omega$ is not bounded a constant is no domination, and besides I still have $f$ around which is only locally integrable, so I can use DC if $M$ is contained in a compact set, otherwise $f$ integrated over $M$ might well be infinite… so can I use DC to simplify things or do I have to use Fatou on the sets of point 4?






      share|cite|improve this answer









      $endgroup$
















        5












        5








        5





        $begingroup$

        I will now outline a proof I managed to come up with, so as to have this outline available online. It seems a web search does not present a proof that doesn't assume continuity on the part of $f$…



        Continuous case



        If $f$ is continuous, the steps are two:




        1. Prove the hypothesis stays true if you substitute smooth compactly supported (henceforth cs) functions with characteristics of balls; this is achieved via smooth transitions, which approximate characteristics of balls pointwise (but evidently not uniformly);

        2. $f$ is continuous, so if $f(x)neq0$ for some $x$, there exists a ball around $x$ where $f$ has constant sign; integrate $f$ times the characteristic of that ball and you should get something nonzero by what I just said, and zero by step 1: a contradiction; so $f=0$ everywhere.


        General case



        The proof here is much longer.





        1. Approximate continuous functions uniformly (and thus in $L^1$ and pointwise a.e.) by convolutions; in other words, define:



          $$rho_n(x)=left{begin{array}{cc}
          n^Ne^{frac{1}{1-|x|^2}} & |x|leq1 \
          0 & text{otherwise}
          end{array}right.,$$



          where $N$ is the dimension of the space $f$ is defined on (i.e., $f:Omegasubseteqmathbb{R}^Ntomathbb{R}$); then let:



          $$rho_nast u(x)=int_{mathrm{supp}(u)}rho_n(x-y)u(y)mathrm{d}y=int_{B(0,frac1n)}rho_n(y)u(x-y)mathrm{d}y.$$



          By the first expression, this is a smooth function; the support is the sum of the ball and the support of $u$, thus if $uin C_c(Omega)$, the support is compact; finally, using the second expression and a few manipulations, one can prove $rho_nast u-u$ has $L^1$ norm less than the supremum of $|tau_yu-u|_{L^1}$ over $|y|<frac1n$, where $tau_yu(x)=u(x-y)$; that $L^1$ norm reduces to integrating over the compact support of $u$, thus $u$ is uniformly continuous there, and this can be used to prove that sup tends to zero, and hence $rho_nast uto u$ in $L^1$; a subsequence thus converges a.e., everything here is bounded by the sup of $u$ which is continuous and compactly supported therefore bounded, so we have a domination and we conclude $frho_nast uto fu$ in $L^1$, so the integrals converge, but the LHS always integrates to 0;



        2. For any $frac1n$, one has a finite cover of a compact set $KsubseteqOmega$ by balls of radius $frac1n$, and thus a partition of unity, i.e. a collection of $phi_i^{(n)}$ which are in bijection with those balls, are each zero outside the matching ball, and sum to 1 on $K$ and less than 1 out of it; plus, they are continuous, so $fsumphi_j^{(n)}$ always integrates to 0; finally, those converge pointwise (at least a.e.) to the indicator of $K$, so since we have a domination we conclude the integrals of $fsumphi_i^{(n)}$ tend to the integral of $f$ over $K$ as $ntoinfty$, but those integrals are all 0 by step 1; hence $f$ integrates to 0 over all compact subsets of $Omega$;


        3. The Lebesgue measure is inner regular, so there is, for any measurable $MsubseteqOmega$, a sequence $K_n$ of compacts such that $mu(K_n)tomu(M)$; this easily implies $L^1$ convergence of the indicators of those compacts to that of $M$; thus, along a subsequence, there is pointwise a.e. convergence; so I should be able to deduce the integrals of $f$ over those compacts converge to that of $f$ over $M$, but I'm not so sure I can use dominate convergence here; perhaps I can just restrict myself to the sets below…;


        4. $f$ is measurable due to integrability, hence $F^+:={f>0},F_-:={f<0}$ are both measurable; their characteristics are approximated by characteristics of compact subsets as shown above; in those cases, I can apply Fatou if Dominated Convergence doesn't apply; Fatou surely yields $int_Omega f^+leq0$, as well as the integral of $f^-$; but those functions are positive, hence those integrals are 0, and they sum to $int_Omega|f|$, which is therefore 0; and bingo: this implies -- at long last!! -- $f=0$ a.e..



        Can someone please verify these steps, and say if I've gone wrong at some point? In particular, in general case step 3, I'm afraid DC is unavailable in general since if $Omega$ is not bounded a constant is no domination, and besides I still have $f$ around which is only locally integrable, so I can use DC if $M$ is contained in a compact set, otherwise $f$ integrated over $M$ might well be infinite… so can I use DC to simplify things or do I have to use Fatou on the sets of point 4?






        share|cite|improve this answer









        $endgroup$



        I will now outline a proof I managed to come up with, so as to have this outline available online. It seems a web search does not present a proof that doesn't assume continuity on the part of $f$…



        Continuous case



        If $f$ is continuous, the steps are two:




        1. Prove the hypothesis stays true if you substitute smooth compactly supported (henceforth cs) functions with characteristics of balls; this is achieved via smooth transitions, which approximate characteristics of balls pointwise (but evidently not uniformly);

        2. $f$ is continuous, so if $f(x)neq0$ for some $x$, there exists a ball around $x$ where $f$ has constant sign; integrate $f$ times the characteristic of that ball and you should get something nonzero by what I just said, and zero by step 1: a contradiction; so $f=0$ everywhere.


        General case



        The proof here is much longer.





        1. Approximate continuous functions uniformly (and thus in $L^1$ and pointwise a.e.) by convolutions; in other words, define:



          $$rho_n(x)=left{begin{array}{cc}
          n^Ne^{frac{1}{1-|x|^2}} & |x|leq1 \
          0 & text{otherwise}
          end{array}right.,$$



          where $N$ is the dimension of the space $f$ is defined on (i.e., $f:Omegasubseteqmathbb{R}^Ntomathbb{R}$); then let:



          $$rho_nast u(x)=int_{mathrm{supp}(u)}rho_n(x-y)u(y)mathrm{d}y=int_{B(0,frac1n)}rho_n(y)u(x-y)mathrm{d}y.$$



          By the first expression, this is a smooth function; the support is the sum of the ball and the support of $u$, thus if $uin C_c(Omega)$, the support is compact; finally, using the second expression and a few manipulations, one can prove $rho_nast u-u$ has $L^1$ norm less than the supremum of $|tau_yu-u|_{L^1}$ over $|y|<frac1n$, where $tau_yu(x)=u(x-y)$; that $L^1$ norm reduces to integrating over the compact support of $u$, thus $u$ is uniformly continuous there, and this can be used to prove that sup tends to zero, and hence $rho_nast uto u$ in $L^1$; a subsequence thus converges a.e., everything here is bounded by the sup of $u$ which is continuous and compactly supported therefore bounded, so we have a domination and we conclude $frho_nast uto fu$ in $L^1$, so the integrals converge, but the LHS always integrates to 0;



        2. For any $frac1n$, one has a finite cover of a compact set $KsubseteqOmega$ by balls of radius $frac1n$, and thus a partition of unity, i.e. a collection of $phi_i^{(n)}$ which are in bijection with those balls, are each zero outside the matching ball, and sum to 1 on $K$ and less than 1 out of it; plus, they are continuous, so $fsumphi_j^{(n)}$ always integrates to 0; finally, those converge pointwise (at least a.e.) to the indicator of $K$, so since we have a domination we conclude the integrals of $fsumphi_i^{(n)}$ tend to the integral of $f$ over $K$ as $ntoinfty$, but those integrals are all 0 by step 1; hence $f$ integrates to 0 over all compact subsets of $Omega$;


        3. The Lebesgue measure is inner regular, so there is, for any measurable $MsubseteqOmega$, a sequence $K_n$ of compacts such that $mu(K_n)tomu(M)$; this easily implies $L^1$ convergence of the indicators of those compacts to that of $M$; thus, along a subsequence, there is pointwise a.e. convergence; so I should be able to deduce the integrals of $f$ over those compacts converge to that of $f$ over $M$, but I'm not so sure I can use dominate convergence here; perhaps I can just restrict myself to the sets below…;


        4. $f$ is measurable due to integrability, hence $F^+:={f>0},F_-:={f<0}$ are both measurable; their characteristics are approximated by characteristics of compact subsets as shown above; in those cases, I can apply Fatou if Dominated Convergence doesn't apply; Fatou surely yields $int_Omega f^+leq0$, as well as the integral of $f^-$; but those functions are positive, hence those integrals are 0, and they sum to $int_Omega|f|$, which is therefore 0; and bingo: this implies -- at long last!! -- $f=0$ a.e..



        Can someone please verify these steps, and say if I've gone wrong at some point? In particular, in general case step 3, I'm afraid DC is unavailable in general since if $Omega$ is not bounded a constant is no domination, and besides I still have $f$ around which is only locally integrable, so I can use DC if $M$ is contained in a compact set, otherwise $f$ integrated over $M$ might well be infinite… so can I use DC to simplify things or do I have to use Fatou on the sets of point 4?







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 26 '15 at 21:32









        MickGMickG

        4,35931956




        4,35931956






























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