Discontinuous function extended as a distribution
$begingroup$
Let $f:mathbb{R}^startomathbb{C}$ be a continuous function on its domain, and suppose : $$exists minmathbb{N}^star,;exists c>0;/;forall xin[-1,1]setminuslbrace0rbrace,;|f(x)|leqslantfrac{c}{|x|^m}$$
The goal is to prove that there exists a distribution $T_finmathcal{D}'(mathbb{R})$ that extends $f$ to $mathbb{R}$, in the sense that if $varphiinmathcal{D}(mathbb{R}^star)$ is a test function whose support does not contain $0$, then $<T_f,varphi>=<f,varphi>$.
Knowing that the principal values are given as : $$<text{vp}left(frac{1}{x}right),varphi>=lim_{varepsilonto0^+}int_{|x|>varepsilon}frac{varphi(x)}{x}text{d}x$$ and : $$<text{vp}left(frac{1}{x^2}right),varphi>=lim_{varepsilonto0^+}left(int_{|x|>varepsilon}frac{varphi(x)}{x^2}text{d}x-frac{2varphi(0)}{varepsilon}right),$$
I would be tempted to look for $T_f$ as something like : $$<T_f,varphi>=lim_{varepsilonto0^+}left(int_{|x|>varepsilon}f(x)varphi(x)text{d}x-beta_varepsilon(varphi)right)$$
Does anyone know something about this ? I can find no litterature, and I have no clue what the $beta_varepsilon(varphi)$ would be besides a sum of coefficients of the form $displaystylefrac{lambdavarphi^{(k)}(0)}{varepsilon^ell}$.
Thanks in advance for your suggestions and pro tips !
real-analysis distribution-theory
asked Dec 3 '18 at 23:16
AnthonyAnthony
628
$endgroup$
|
show 1 more comment
$begingroup$
Let $f:mathbb{R}^startomathbb{C}$ be a continuous function on its domain, and suppose : $$exists minmathbb{N}^star,;exists c>0;/;forall xin[-1,1]setminuslbrace0rbrace,;|f(x)|leqslantfrac{c}{|x|^m}$$
The goal is to prove that there exists a distribution $T_finmathcal{D}'(mathbb{R})$ that extends $f$ to $mathbb{R}$, in the sense that if $varphiinmathcal{D}(mathbb{R}^star)$ is a test function whose support does not contain $0$, then $<T_f,varphi>=<f,varphi>$.
Knowing that the principal values are given as : $$<text{vp}left(frac{1}{x}right),varphi>=lim_{varepsilonto0^+}int_{|x|>varepsilon}frac{varphi(x)}{x}text{d}x$$ and : $$<text{vp}left(frac{1}{x^2}right),varphi>=lim_{varepsilonto0^+}left(int_{|x|>varepsilon}frac{varphi(x)}{x^2}text{d}x-frac{2varphi(0)}{varepsilon}right),$$
I would be tempted to look for $T_f$ as something like : $$<T_f,varphi>=lim_{varepsilonto0^+}left(int_{|x|>varepsilon}f(x)varphi(x)text{d}x-beta_varepsilon(varphi)right)$$
Does anyone know something about this ? I can find no litterature, and I have no clue what the $beta_varepsilon(varphi)$ would be besides a sum of coefficients of the form $displaystylefrac{lambdavarphi^{(k)}(0)}{varepsilon^ell}$.
Thanks in advance for your suggestions and pro tips !
real-analysis distribution-theory
asked Dec 3 '18 at 23:16
AnthonyAnthony
628
$endgroup$
1
$begingroup$
Look at $T_{f,psi,M}(varphi) = langle f, varphi - psi sum_{k=0}^M frac{varphi^{(k)}(0)}{k!} x^krangle$ where you have fixed some $psi in C^infty_c$ constant on $[-a,a]$. The principal value of $1/x$ is obtained with $M=0$ and $psi$ even.
$endgroup$
– reuns
Dec 3 '18 at 23:20
$begingroup$
@reuns Is the goal to try to take $ato0^+$ afterwards ? :)
$endgroup$
– Anthony
Dec 3 '18 at 23:36
$begingroup$
No ${}{}{}{}{}$
$endgroup$
– reuns
Dec 3 '18 at 23:38
$begingroup$
@reuns okay I get it for vp(1/x) ! For vp(1/x²), I need to take $M=1$ and $psi$ even as well, but I don't know what to ask more... Btw, $psi$ has to be constant equal to $1$ on a neighboor of $0$ right ? Do I have to make $ato+infty$ ?! :O
$endgroup$
– Anthony
Dec 4 '18 at 0:34
1
$begingroup$
?? Why don't you want to keep $a$ fixed finite non-zero ? Do you see why $varphi mapsto varphi - psi sum_{k=0}^M frac{varphi^{(k)}(0)}{k!} x^k$ is a continuous linear map $C^infty_c to C^infty_c$ ? Thus composing with distributions (continuous linear forms) works well. Yes $psi$ is constant $=1$ on $[a,a]$ (or at least $psi(0) = 1,psi^{(k)}(0) = 0$ for $1 le k le M$) and $M ge m-1$. I would ask what are the distributional derivatives of $T_{1/x,psi,M}$, to obtain a formula for all the derivatives of $vp(1/x)$.
$endgroup$
– reuns
Dec 4 '18 at 0:39
|
show 1 more comment
$begingroup$
Let $f:mathbb{R}^startomathbb{C}$ be a continuous function on its domain, and suppose : $$exists minmathbb{N}^star,;exists c>0;/;forall xin[-1,1]setminuslbrace0rbrace,;|f(x)|leqslantfrac{c}{|x|^m}$$
The goal is to prove that there exists a distribution $T_finmathcal{D}'(mathbb{R})$ that extends $f$ to $mathbb{R}$, in the sense that if $varphiinmathcal{D}(mathbb{R}^star)$ is a test function whose support does not contain $0$, then $<T_f,varphi>=<f,varphi>$.
Knowing that the principal values are given as : $$<text{vp}left(frac{1}{x}right),varphi>=lim_{varepsilonto0^+}int_{|x|>varepsilon}frac{varphi(x)}{x}text{d}x$$ and : $$<text{vp}left(frac{1}{x^2}right),varphi>=lim_{varepsilonto0^+}left(int_{|x|>varepsilon}frac{varphi(x)}{x^2}text{d}x-frac{2varphi(0)}{varepsilon}right),$$
I would be tempted to look for $T_f$ as something like : $$<T_f,varphi>=lim_{varepsilonto0^+}left(int_{|x|>varepsilon}f(x)varphi(x)text{d}x-beta_varepsilon(varphi)right)$$
Does anyone know something about this ? I can find no litterature, and I have no clue what the $beta_varepsilon(varphi)$ would be besides a sum of coefficients of the form $displaystylefrac{lambdavarphi^{(k)}(0)}{varepsilon^ell}$.
Thanks in advance for your suggestions and pro tips !
real-analysis distribution-theory
asked Dec 3 '18 at 23:16
AnthonyAnthony
628
$endgroup$
Let $f:mathbb{R}^startomathbb{C}$ be a continuous function on its domain, and suppose : $$exists minmathbb{N}^star,;exists c>0;/;forall xin[-1,1]setminuslbrace0rbrace,;|f(x)|leqslantfrac{c}{|x|^m}$$
The goal is to prove that there exists a distribution $T_finmathcal{D}'(mathbb{R})$ that extends $f$ to $mathbb{R}$, in the sense that if $varphiinmathcal{D}(mathbb{R}^star)$ is a test function whose support does not contain $0$, then $<T_f,varphi>=<f,varphi>$.
Knowing that the principal values are given as : $$<text{vp}left(frac{1}{x}right),varphi>=lim_{varepsilonto0^+}int_{|x|>varepsilon}frac{varphi(x)}{x}text{d}x$$ and : $$<text{vp}left(frac{1}{x^2}right),varphi>=lim_{varepsilonto0^+}left(int_{|x|>varepsilon}frac{varphi(x)}{x^2}text{d}x-frac{2varphi(0)}{varepsilon}right),$$
I would be tempted to look for $T_f$ as something like : $$<T_f,varphi>=lim_{varepsilonto0^+}left(int_{|x|>varepsilon}f(x)varphi(x)text{d}x-beta_varepsilon(varphi)right)$$
Does anyone know something about this ? I can find no litterature, and I have no clue what the $beta_varepsilon(varphi)$ would be besides a sum of coefficients of the form $displaystylefrac{lambdavarphi^{(k)}(0)}{varepsilon^ell}$.
Thanks in advance for your suggestions and pro tips !
real-analysis distribution-theory
real-analysis distribution-theory
asked Dec 3 '18 at 23:16
AnthonyAnthony
628
asked Dec 3 '18 at 23:16
AnthonyAnthony
628
asked Dec 3 '18 at 23:16
AnthonyAnthony
628
asked Dec 3 '18 at 23:16
AnthonyAnthony
628
asked Dec 3 '18 at 23:16
AnthonyAnthony
628
628
1
$begingroup$
Look at $T_{f,psi,M}(varphi) = langle f, varphi - psi sum_{k=0}^M frac{varphi^{(k)}(0)}{k!} x^krangle$ where you have fixed some $psi in C^infty_c$ constant on $[-a,a]$. The principal value of $1/x$ is obtained with $M=0$ and $psi$ even.
$endgroup$
– reuns
Dec 3 '18 at 23:20
$begingroup$
@reuns Is the goal to try to take $ato0^+$ afterwards ? :)
$endgroup$
– Anthony
Dec 3 '18 at 23:36
$begingroup$
No ${}{}{}{}{}$
$endgroup$
– reuns
Dec 3 '18 at 23:38
$begingroup$
@reuns okay I get it for vp(1/x) ! For vp(1/x²), I need to take $M=1$ and $psi$ even as well, but I don't know what to ask more... Btw, $psi$ has to be constant equal to $1$ on a neighboor of $0$ right ? Do I have to make $ato+infty$ ?! :O
$endgroup$
– Anthony
Dec 4 '18 at 0:34
1
$begingroup$
?? Why don't you want to keep $a$ fixed finite non-zero ? Do you see why $varphi mapsto varphi - psi sum_{k=0}^M frac{varphi^{(k)}(0)}{k!} x^k$ is a continuous linear map $C^infty_c to C^infty_c$ ? Thus composing with distributions (continuous linear forms) works well. Yes $psi$ is constant $=1$ on $[a,a]$ (or at least $psi(0) = 1,psi^{(k)}(0) = 0$ for $1 le k le M$) and $M ge m-1$. I would ask what are the distributional derivatives of $T_{1/x,psi,M}$, to obtain a formula for all the derivatives of $vp(1/x)$.
$endgroup$
– reuns
Dec 4 '18 at 0:39
|
show 1 more comment
1
$begingroup$
Look at $T_{f,psi,M}(varphi) = langle f, varphi - psi sum_{k=0}^M frac{varphi^{(k)}(0)}{k!} x^krangle$ where you have fixed some $psi in C^infty_c$ constant on $[-a,a]$. The principal value of $1/x$ is obtained with $M=0$ and $psi$ even.
$endgroup$
– reuns
Dec 3 '18 at 23:20
$begingroup$
@reuns Is the goal to try to take $ato0^+$ afterwards ? :)
$endgroup$
– Anthony
Dec 3 '18 at 23:36
$begingroup$
No ${}{}{}{}{}$
$endgroup$
– reuns
Dec 3 '18 at 23:38
$begingroup$
@reuns okay I get it for vp(1/x) ! For vp(1/x²), I need to take $M=1$ and $psi$ even as well, but I don't know what to ask more... Btw, $psi$ has to be constant equal to $1$ on a neighboor of $0$ right ? Do I have to make $ato+infty$ ?! :O
$endgroup$
– Anthony
Dec 4 '18 at 0:34
1
$begingroup$
?? Why don't you want to keep $a$ fixed finite non-zero ? Do you see why $varphi mapsto varphi - psi sum_{k=0}^M frac{varphi^{(k)}(0)}{k!} x^k$ is a continuous linear map $C^infty_c to C^infty_c$ ? Thus composing with distributions (continuous linear forms) works well. Yes $psi$ is constant $=1$ on $[a,a]$ (or at least $psi(0) = 1,psi^{(k)}(0) = 0$ for $1 le k le M$) and $M ge m-1$. I would ask what are the distributional derivatives of $T_{1/x,psi,M}$, to obtain a formula for all the derivatives of $vp(1/x)$.
$endgroup$
– reuns
Dec 4 '18 at 0:39
1
1
$begingroup$
Look at $T_{f,psi,M}(varphi) = langle f, varphi - psi sum_{k=0}^M frac{varphi^{(k)}(0)}{k!} x^krangle$ where you have fixed some $psi in C^infty_c$ constant on $[-a,a]$. The principal value of $1/x$ is obtained with $M=0$ and $psi$ even.
$endgroup$
– reuns
Dec 3 '18 at 23:20
$begingroup$
Look at $T_{f,psi,M}(varphi) = langle f, varphi - psi sum_{k=0}^M frac{varphi^{(k)}(0)}{k!} x^krangle$ where you have fixed some $psi in C^infty_c$ constant on $[-a,a]$. The principal value of $1/x$ is obtained with $M=0$ and $psi$ even.
$endgroup$
– reuns
Dec 3 '18 at 23:20
$begingroup$
@reuns Is the goal to try to take $ato0^+$ afterwards ? :)
$endgroup$
– Anthony
Dec 3 '18 at 23:36
$begingroup$
@reuns Is the goal to try to take $ato0^+$ afterwards ? :)
$endgroup$
– Anthony
Dec 3 '18 at 23:36
$begingroup$
No ${}{}{}{}{}$
$endgroup$
– reuns
Dec 3 '18 at 23:38
$begingroup$
No ${}{}{}{}{}$
$endgroup$
– reuns
Dec 3 '18 at 23:38
$begingroup$
@reuns okay I get it for vp(1/x) ! For vp(1/x²), I need to take $M=1$ and $psi$ even as well, but I don't know what to ask more... Btw, $psi$ has to be constant equal to $1$ on a neighboor of $0$ right ? Do I have to make $ato+infty$ ?! :O
$endgroup$
– Anthony
Dec 4 '18 at 0:34
$begingroup$
@reuns okay I get it for vp(1/x) ! For vp(1/x²), I need to take $M=1$ and $psi$ even as well, but I don't know what to ask more... Btw, $psi$ has to be constant equal to $1$ on a neighboor of $0$ right ? Do I have to make $ato+infty$ ?! :O
$endgroup$
– Anthony
Dec 4 '18 at 0:34
1
1
$begingroup$
?? Why don't you want to keep $a$ fixed finite non-zero ? Do you see why $varphi mapsto varphi - psi sum_{k=0}^M frac{varphi^{(k)}(0)}{k!} x^k$ is a continuous linear map $C^infty_c to C^infty_c$ ? Thus composing with distributions (continuous linear forms) works well. Yes $psi$ is constant $=1$ on $[a,a]$ (or at least $psi(0) = 1,psi^{(k)}(0) = 0$ for $1 le k le M$) and $M ge m-1$. I would ask what are the distributional derivatives of $T_{1/x,psi,M}$, to obtain a formula for all the derivatives of $vp(1/x)$.
$endgroup$
– reuns
Dec 4 '18 at 0:39
$begingroup$
?? Why don't you want to keep $a$ fixed finite non-zero ? Do you see why $varphi mapsto varphi - psi sum_{k=0}^M frac{varphi^{(k)}(0)}{k!} x^k$ is a continuous linear map $C^infty_c to C^infty_c$ ? Thus composing with distributions (continuous linear forms) works well. Yes $psi$ is constant $=1$ on $[a,a]$ (or at least $psi(0) = 1,psi^{(k)}(0) = 0$ for $1 le k le M$) and $M ge m-1$. I would ask what are the distributional derivatives of $T_{1/x,psi,M}$, to obtain a formula for all the derivatives of $vp(1/x)$.
$endgroup$
– reuns
Dec 4 '18 at 0:39
|
show 1 more comment
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1
$begingroup$
Look at $T_{f,psi,M}(varphi) = langle f, varphi - psi sum_{k=0}^M frac{varphi^{(k)}(0)}{k!} x^krangle$ where you have fixed some $psi in C^infty_c$ constant on $[-a,a]$. The principal value of $1/x$ is obtained with $M=0$ and $psi$ even.
$endgroup$
– reuns
Dec 3 '18 at 23:20
$begingroup$
@reuns Is the goal to try to take $ato0^+$ afterwards ? :)
$endgroup$
– Anthony
Dec 3 '18 at 23:36
$begingroup$
No ${}{}{}{}{}$
$endgroup$
– reuns
Dec 3 '18 at 23:38
$begingroup$
@reuns okay I get it for vp(1/x) ! For vp(1/x²), I need to take $M=1$ and $psi$ even as well, but I don't know what to ask more... Btw, $psi$ has to be constant equal to $1$ on a neighboor of $0$ right ? Do I have to make $ato+infty$ ?! :O
$endgroup$
– Anthony
Dec 4 '18 at 0:34
1
$begingroup$
?? Why don't you want to keep $a$ fixed finite non-zero ? Do you see why $varphi mapsto varphi - psi sum_{k=0}^M frac{varphi^{(k)}(0)}{k!} x^k$ is a continuous linear map $C^infty_c to C^infty_c$ ? Thus composing with distributions (continuous linear forms) works well. Yes $psi$ is constant $=1$ on $[a,a]$ (or at least $psi(0) = 1,psi^{(k)}(0) = 0$ for $1 le k le M$) and $M ge m-1$. I would ask what are the distributional derivatives of $T_{1/x,psi,M}$, to obtain a formula for all the derivatives of $vp(1/x)$.
$endgroup$
– reuns
Dec 4 '18 at 0:39