If $phi: [0,, 1] to X $ be a sequentially weakly continuous, then $f(t) = |φ(t)| $ is bounded
Let $X$ be a Banach space. Let $phi: [0,, 1] to X $ be a sequentially weakly continuous function, that is,
$$forall ,(t_n) subset [0, 1],,,, t_n to t Rightarrow phi(t_n) rightharpoonup phi(t).$$
Consider the function $ f : [0, 1] to mathbb{R} $ defined by $f(t) = |φ(t)| $ for every $t in [0,, 1]$. Prove that
$f$ is bounded,
there exists the minimum of $f$ on $[0,, 1]$.
My solution.
Since $|x|geq 0$, the function $f$ is bounded from below and the infimum $, I:=inf_{[0,1]} f ,,$ is finite. Now we apply the direct method in calculus of variations.
Let $t_n$ be a minimizing sequence, that is $f(t_n)to I$. Since $[0, ,1]$ is compact there exists a subsequence $t_{n_k}$ converging to $overline{t}in [0,,1]$. The norm is weakly lower semicontinuous so $f$ is semicontinuous and in particular
$$f(overline t) leq liminf_{ktoinfty}f(t_{n_k})=lim_{ntoinfty}f(t_{n})=I,$$
but by definition of infimum $Ileq f(overline t)$ and we get $$f(overline t)=I.$$
Questions:
It is necessary "$X$ is a Banach space" ? I think the answer is no.
It is possible to prove that $f$ is upper bounded ? The exercise say "Prove that $f$ is bounded" but I think it's a mistake. I think the author wanted to write "Prove that $f$ is bounded from below" (indeed this is enough for the existence of the minimum).
Thanks in advance.
proof-verification alternative-proof calculus-of-variations weak-convergence
asked Jun 30 at 21:47
Ef_Ci
1458
|
show 1 more comment
Let $X$ be a Banach space. Let $phi: [0,, 1] to X $ be a sequentially weakly continuous function, that is,
$$forall ,(t_n) subset [0, 1],,,, t_n to t Rightarrow phi(t_n) rightharpoonup phi(t).$$
Consider the function $ f : [0, 1] to mathbb{R} $ defined by $f(t) = |φ(t)| $ for every $t in [0,, 1]$. Prove that
$f$ is bounded,
there exists the minimum of $f$ on $[0,, 1]$.
My solution.
Since $|x|geq 0$, the function $f$ is bounded from below and the infimum $, I:=inf_{[0,1]} f ,,$ is finite. Now we apply the direct method in calculus of variations.
Let $t_n$ be a minimizing sequence, that is $f(t_n)to I$. Since $[0, ,1]$ is compact there exists a subsequence $t_{n_k}$ converging to $overline{t}in [0,,1]$. The norm is weakly lower semicontinuous so $f$ is semicontinuous and in particular
$$f(overline t) leq liminf_{ktoinfty}f(t_{n_k})=lim_{ntoinfty}f(t_{n})=I,$$
but by definition of infimum $Ileq f(overline t)$ and we get $$f(overline t)=I.$$
Questions:
It is necessary "$X$ is a Banach space" ? I think the answer is no.
It is possible to prove that $f$ is upper bounded ? The exercise say "Prove that $f$ is bounded" but I think it's a mistake. I think the author wanted to write "Prove that $f$ is bounded from below" (indeed this is enough for the existence of the minimum).
Thanks in advance.
proof-verification alternative-proof calculus-of-variations weak-convergence
asked Jun 30 at 21:47
Ef_Ci
1458
Since you are dealing with metric spaces, $varphi$ is continuous, hence $f$ too and since $[0, 1]$ is compact, the minimiser exists & $f$ is bounded.
– Will M.
Jun 30 at 21:50
The norm is only semicontinuous. Anyway, this is not a problem. I have proved the existence of the minimizer. The main question that remains open is to prove if $f$ it's also upper bounded.
– Ef_Ci
Jun 30 at 21:53
The norm $x mapsto |x|$ from $X$ into $Bbb R$ is continuous.
– Will M.
Jun 30 at 21:55
Yes, but with respect to the strong convergence and it is only lower semicontinuous with respect to the weak convergence: math.stackexchange.com/questions/470217/… Anyway, I underline that my question is about the boundedness.
– Ef_Ci
Jun 30 at 22:01
In that case, you need to state the exercise so that one that is not thinking what you are thinking exactly can understand what you mean. The exercise is ambiguous as stated.
– Will M.
Jun 30 at 22:03
|
show 1 more comment
Let $X$ be a Banach space. Let $phi: [0,, 1] to X $ be a sequentially weakly continuous function, that is,
$$forall ,(t_n) subset [0, 1],,,, t_n to t Rightarrow phi(t_n) rightharpoonup phi(t).$$
Consider the function $ f : [0, 1] to mathbb{R} $ defined by $f(t) = |φ(t)| $ for every $t in [0,, 1]$. Prove that
$f$ is bounded,
there exists the minimum of $f$ on $[0,, 1]$.
My solution.
Since $|x|geq 0$, the function $f$ is bounded from below and the infimum $, I:=inf_{[0,1]} f ,,$ is finite. Now we apply the direct method in calculus of variations.
Let $t_n$ be a minimizing sequence, that is $f(t_n)to I$. Since $[0, ,1]$ is compact there exists a subsequence $t_{n_k}$ converging to $overline{t}in [0,,1]$. The norm is weakly lower semicontinuous so $f$ is semicontinuous and in particular
$$f(overline t) leq liminf_{ktoinfty}f(t_{n_k})=lim_{ntoinfty}f(t_{n})=I,$$
but by definition of infimum $Ileq f(overline t)$ and we get $$f(overline t)=I.$$
Questions:
It is necessary "$X$ is a Banach space" ? I think the answer is no.
It is possible to prove that $f$ is upper bounded ? The exercise say "Prove that $f$ is bounded" but I think it's a mistake. I think the author wanted to write "Prove that $f$ is bounded from below" (indeed this is enough for the existence of the minimum).
Thanks in advance.
proof-verification alternative-proof calculus-of-variations weak-convergence
asked Jun 30 at 21:47
Ef_Ci
1458
Let $X$ be a Banach space. Let $phi: [0,, 1] to X $ be a sequentially weakly continuous function, that is,
$$forall ,(t_n) subset [0, 1],,,, t_n to t Rightarrow phi(t_n) rightharpoonup phi(t).$$
Consider the function $ f : [0, 1] to mathbb{R} $ defined by $f(t) = |φ(t)| $ for every $t in [0,, 1]$. Prove that
$f$ is bounded,
there exists the minimum of $f$ on $[0,, 1]$.
My solution.
Since $|x|geq 0$, the function $f$ is bounded from below and the infimum $, I:=inf_{[0,1]} f ,,$ is finite. Now we apply the direct method in calculus of variations.
Let $t_n$ be a minimizing sequence, that is $f(t_n)to I$. Since $[0, ,1]$ is compact there exists a subsequence $t_{n_k}$ converging to $overline{t}in [0,,1]$. The norm is weakly lower semicontinuous so $f$ is semicontinuous and in particular
$$f(overline t) leq liminf_{ktoinfty}f(t_{n_k})=lim_{ntoinfty}f(t_{n})=I,$$
but by definition of infimum $Ileq f(overline t)$ and we get $$f(overline t)=I.$$
Questions:
It is necessary "$X$ is a Banach space" ? I think the answer is no.
It is possible to prove that $f$ is upper bounded ? The exercise say "Prove that $f$ is bounded" but I think it's a mistake. I think the author wanted to write "Prove that $f$ is bounded from below" (indeed this is enough for the existence of the minimum).
Thanks in advance.
proof-verification alternative-proof calculus-of-variations weak-convergence
proof-verification alternative-proof calculus-of-variations weak-convergence
asked Jun 30 at 21:47
Ef_Ci
1458
asked Jun 30 at 21:47
Ef_Ci
1458
asked Jun 30 at 21:47
Ef_Ci
1458
asked Jun 30 at 21:47
Ef_Ci
1458
asked Jun 30 at 21:47
Ef_Ci
1458
1458
Since you are dealing with metric spaces, $varphi$ is continuous, hence $f$ too and since $[0, 1]$ is compact, the minimiser exists & $f$ is bounded.
– Will M.
Jun 30 at 21:50
The norm is only semicontinuous. Anyway, this is not a problem. I have proved the existence of the minimizer. The main question that remains open is to prove if $f$ it's also upper bounded.
– Ef_Ci
Jun 30 at 21:53
The norm $x mapsto |x|$ from $X$ into $Bbb R$ is continuous.
– Will M.
Jun 30 at 21:55
Yes, but with respect to the strong convergence and it is only lower semicontinuous with respect to the weak convergence: math.stackexchange.com/questions/470217/… Anyway, I underline that my question is about the boundedness.
– Ef_Ci
Jun 30 at 22:01
In that case, you need to state the exercise so that one that is not thinking what you are thinking exactly can understand what you mean. The exercise is ambiguous as stated.
– Will M.
Jun 30 at 22:03
|
show 1 more comment
Since you are dealing with metric spaces, $varphi$ is continuous, hence $f$ too and since $[0, 1]$ is compact, the minimiser exists & $f$ is bounded.
– Will M.
Jun 30 at 21:50
The norm is only semicontinuous. Anyway, this is not a problem. I have proved the existence of the minimizer. The main question that remains open is to prove if $f$ it's also upper bounded.
– Ef_Ci
Jun 30 at 21:53
The norm $x mapsto |x|$ from $X$ into $Bbb R$ is continuous.
– Will M.
Jun 30 at 21:55
Yes, but with respect to the strong convergence and it is only lower semicontinuous with respect to the weak convergence: math.stackexchange.com/questions/470217/… Anyway, I underline that my question is about the boundedness.
– Ef_Ci
Jun 30 at 22:01
In that case, you need to state the exercise so that one that is not thinking what you are thinking exactly can understand what you mean. The exercise is ambiguous as stated.
– Will M.
Jun 30 at 22:03
Since you are dealing with metric spaces, $varphi$ is continuous, hence $f$ too and since $[0, 1]$ is compact, the minimiser exists & $f$ is bounded.
– Will M.
Jun 30 at 21:50
Since you are dealing with metric spaces, $varphi$ is continuous, hence $f$ too and since $[0, 1]$ is compact, the minimiser exists & $f$ is bounded.
– Will M.
Jun 30 at 21:50
The norm is only semicontinuous. Anyway, this is not a problem. I have proved the existence of the minimizer. The main question that remains open is to prove if $f$ it's also upper bounded.
– Ef_Ci
Jun 30 at 21:53
The norm is only semicontinuous. Anyway, this is not a problem. I have proved the existence of the minimizer. The main question that remains open is to prove if $f$ it's also upper bounded.
– Ef_Ci
Jun 30 at 21:53
The norm $x mapsto |x|$ from $X$ into $Bbb R$ is continuous.
– Will M.
Jun 30 at 21:55
The norm $x mapsto |x|$ from $X$ into $Bbb R$ is continuous.
– Will M.
Jun 30 at 21:55
Yes, but with respect to the strong convergence and it is only lower semicontinuous with respect to the weak convergence: math.stackexchange.com/questions/470217/… Anyway, I underline that my question is about the boundedness.
– Ef_Ci
Jun 30 at 22:01
Yes, but with respect to the strong convergence and it is only lower semicontinuous with respect to the weak convergence: math.stackexchange.com/questions/470217/… Anyway, I underline that my question is about the boundedness.
– Ef_Ci
Jun 30 at 22:01
In that case, you need to state the exercise so that one that is not thinking what you are thinking exactly can understand what you mean. The exercise is ambiguous as stated.
– Will M.
Jun 30 at 22:03
In that case, you need to state the exercise so that one that is not thinking what you are thinking exactly can understand what you mean. The exercise is ambiguous as stated.
– Will M.
Jun 30 at 22:03
|
show 1 more comment
2 Answers
2
active
oldest
votes
If this is not a solution, then the exercise as written makes no sense.
Assume, for the sake of reaching an absurd, that $f$ is unbounded. There would exist a sequence $t_n to t$ such that $f(t_n) > n,$ but since $varphi(t_n) to varphi(t)$ it turns out $varphi(t_n)$ is bounded (which means $|varphi(t_n)|$ is a bounded sequence of real numbers), hence $f$ is bounded as well, an absurd.
answered Jun 30 at 22:27
Will M.
2,377314
To be honest, the "weakly" part confused me because "weak topologies" are defined usually on the dual of a space.
– Will M.
Jun 30 at 22:41
Thanks. It remains a question: "It is necessary "X is a Banach space" ? I think the answer is no."
– Ef_Ci
Jun 30 at 22:42
add a comment |
The answer by Will M. entirely glosses over the essential part. It is true that every weakly bounded sequence is norm bounded, yet this is in no way obvious, but a consequence of the uniform boundedness principle.
To summarize: Since $phi$ is weakly continuous, its image is weakly compact and thus weakly bounded. By the uniform boundedness principle the image is also norm bounded.
answered Nov 26 at 10:06
MaoWao
2,333616
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If this is not a solution, then the exercise as written makes no sense.
Assume, for the sake of reaching an absurd, that $f$ is unbounded. There would exist a sequence $t_n to t$ such that $f(t_n) > n,$ but since $varphi(t_n) to varphi(t)$ it turns out $varphi(t_n)$ is bounded (which means $|varphi(t_n)|$ is a bounded sequence of real numbers), hence $f$ is bounded as well, an absurd.
answered Jun 30 at 22:27
Will M.
2,377314
To be honest, the "weakly" part confused me because "weak topologies" are defined usually on the dual of a space.
– Will M.
Jun 30 at 22:41
Thanks. It remains a question: "It is necessary "X is a Banach space" ? I think the answer is no."
– Ef_Ci
Jun 30 at 22:42
add a comment |
If this is not a solution, then the exercise as written makes no sense.
Assume, for the sake of reaching an absurd, that $f$ is unbounded. There would exist a sequence $t_n to t$ such that $f(t_n) > n,$ but since $varphi(t_n) to varphi(t)$ it turns out $varphi(t_n)$ is bounded (which means $|varphi(t_n)|$ is a bounded sequence of real numbers), hence $f$ is bounded as well, an absurd.
answered Jun 30 at 22:27
Will M.
2,377314
To be honest, the "weakly" part confused me because "weak topologies" are defined usually on the dual of a space.
– Will M.
Jun 30 at 22:41
Thanks. It remains a question: "It is necessary "X is a Banach space" ? I think the answer is no."
– Ef_Ci
Jun 30 at 22:42
add a comment |
If this is not a solution, then the exercise as written makes no sense.
Assume, for the sake of reaching an absurd, that $f$ is unbounded. There would exist a sequence $t_n to t$ such that $f(t_n) > n,$ but since $varphi(t_n) to varphi(t)$ it turns out $varphi(t_n)$ is bounded (which means $|varphi(t_n)|$ is a bounded sequence of real numbers), hence $f$ is bounded as well, an absurd.
answered Jun 30 at 22:27
Will M.
2,377314
If this is not a solution, then the exercise as written makes no sense.
Assume, for the sake of reaching an absurd, that $f$ is unbounded. There would exist a sequence $t_n to t$ such that $f(t_n) > n,$ but since $varphi(t_n) to varphi(t)$ it turns out $varphi(t_n)$ is bounded (which means $|varphi(t_n)|$ is a bounded sequence of real numbers), hence $f$ is bounded as well, an absurd.
answered Jun 30 at 22:27
Will M.
2,377314
answered Jun 30 at 22:27
Will M.
2,377314
answered Jun 30 at 22:27
Will M.
2,377314
answered Jun 30 at 22:27
Will M.
2,377314
2,377314
To be honest, the "weakly" part confused me because "weak topologies" are defined usually on the dual of a space.
– Will M.
Jun 30 at 22:41
Thanks. It remains a question: "It is necessary "X is a Banach space" ? I think the answer is no."
– Ef_Ci
Jun 30 at 22:42
add a comment |
To be honest, the "weakly" part confused me because "weak topologies" are defined usually on the dual of a space.
– Will M.
Jun 30 at 22:41
Thanks. It remains a question: "It is necessary "X is a Banach space" ? I think the answer is no."
– Ef_Ci
Jun 30 at 22:42
To be honest, the "weakly" part confused me because "weak topologies" are defined usually on the dual of a space.
– Will M.
Jun 30 at 22:41
To be honest, the "weakly" part confused me because "weak topologies" are defined usually on the dual of a space.
– Will M.
Jun 30 at 22:41
Thanks. It remains a question: "It is necessary "X is a Banach space" ? I think the answer is no."
– Ef_Ci
Jun 30 at 22:42
Thanks. It remains a question: "It is necessary "X is a Banach space" ? I think the answer is no."
– Ef_Ci
Jun 30 at 22:42
add a comment |
The answer by Will M. entirely glosses over the essential part. It is true that every weakly bounded sequence is norm bounded, yet this is in no way obvious, but a consequence of the uniform boundedness principle.
To summarize: Since $phi$ is weakly continuous, its image is weakly compact and thus weakly bounded. By the uniform boundedness principle the image is also norm bounded.
answered Nov 26 at 10:06
MaoWao
2,333616
add a comment |
The answer by Will M. entirely glosses over the essential part. It is true that every weakly bounded sequence is norm bounded, yet this is in no way obvious, but a consequence of the uniform boundedness principle.
To summarize: Since $phi$ is weakly continuous, its image is weakly compact and thus weakly bounded. By the uniform boundedness principle the image is also norm bounded.
answered Nov 26 at 10:06
MaoWao
2,333616
add a comment |
The answer by Will M. entirely glosses over the essential part. It is true that every weakly bounded sequence is norm bounded, yet this is in no way obvious, but a consequence of the uniform boundedness principle.
To summarize: Since $phi$ is weakly continuous, its image is weakly compact and thus weakly bounded. By the uniform boundedness principle the image is also norm bounded.
answered Nov 26 at 10:06
MaoWao
2,333616
The answer by Will M. entirely glosses over the essential part. It is true that every weakly bounded sequence is norm bounded, yet this is in no way obvious, but a consequence of the uniform boundedness principle.
To summarize: Since $phi$ is weakly continuous, its image is weakly compact and thus weakly bounded. By the uniform boundedness principle the image is also norm bounded.
answered Nov 26 at 10:06
MaoWao
2,333616
answered Nov 26 at 10:06
MaoWao
2,333616
answered Nov 26 at 10:06
MaoWao
2,333616
answered Nov 26 at 10:06
MaoWao
2,333616
2,333616
add a comment |
add a comment |
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Since you are dealing with metric spaces, $varphi$ is continuous, hence $f$ too and since $[0, 1]$ is compact, the minimiser exists & $f$ is bounded.
– Will M.
Jun 30 at 21:50
The norm is only semicontinuous. Anyway, this is not a problem. I have proved the existence of the minimizer. The main question that remains open is to prove if $f$ it's also upper bounded.
– Ef_Ci
Jun 30 at 21:53
The norm $x mapsto |x|$ from $X$ into $Bbb R$ is continuous.
– Will M.
Jun 30 at 21:55
Yes, but with respect to the strong convergence and it is only lower semicontinuous with respect to the weak convergence: math.stackexchange.com/questions/470217/… Anyway, I underline that my question is about the boundedness.
– Ef_Ci
Jun 30 at 22:01
In that case, you need to state the exercise so that one that is not thinking what you are thinking exactly can understand what you mean. The exercise is ambiguous as stated.
– Will M.
Jun 30 at 22:03