Convolution of a compactly supported function and an $L^1$ function.












2














I have these related questions here that I could really use some help on. I believe there is a related question here although I don't think it is exactly the same...




1) Let $fin L^1(mathbb R)$ and $varphi$ a smooth compactly
supported function. Prove that $fast varphi$ is a smooth function
whose derivative is $fast varphi'$.




Any help would be much appreciated. I have been trying to work on this for the past couple days but all my ideas didn't lead me very far. Namely, I was trying to use the definition of a convolution. I am not sure though how the smooth and compactly supported $L^1$ function fits into this...






real-analysis functional-analysis hilbert-spaces convolution







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  • 1




    For the first part, use the definition of the convolution (and maybe the identity $f ast g = g ast f$), and then differentiate under the integral sign (you need to show that this is allowed).
    – PhoemueX
    Dec 7 at 8:47










  • The direct approach is to find a bound for $f ast varphi' - f ast (varphi(.+y)-varphi)/y$. The less direct one is to look at $int f ast varphi'$
    – reuns
    Dec 7 at 16:44
















2














I have these related questions here that I could really use some help on. I believe there is a related question here although I don't think it is exactly the same...




1) Let $fin L^1(mathbb R)$ and $varphi$ a smooth compactly
supported function. Prove that $fast varphi$ is a smooth function
whose derivative is $fast varphi'$.




Any help would be much appreciated. I have been trying to work on this for the past couple days but all my ideas didn't lead me very far. Namely, I was trying to use the definition of a convolution. I am not sure though how the smooth and compactly supported $L^1$ function fits into this...






real-analysis functional-analysis hilbert-spaces convolution







share|cite|improve this question




















  • 1




    For the first part, use the definition of the convolution (and maybe the identity $f ast g = g ast f$), and then differentiate under the integral sign (you need to show that this is allowed).
    – PhoemueX
    Dec 7 at 8:47










  • The direct approach is to find a bound for $f ast varphi' - f ast (varphi(.+y)-varphi)/y$. The less direct one is to look at $int f ast varphi'$
    – reuns
    Dec 7 at 16:44














2












2








2







I have these related questions here that I could really use some help on. I believe there is a related question here although I don't think it is exactly the same...




1) Let $fin L^1(mathbb R)$ and $varphi$ a smooth compactly
supported function. Prove that $fast varphi$ is a smooth function
whose derivative is $fast varphi'$.




Any help would be much appreciated. I have been trying to work on this for the past couple days but all my ideas didn't lead me very far. Namely, I was trying to use the definition of a convolution. I am not sure though how the smooth and compactly supported $L^1$ function fits into this...






real-analysis functional-analysis hilbert-spaces convolution







share|cite|improve this question















I have these related questions here that I could really use some help on. I believe there is a related question here although I don't think it is exactly the same...




1) Let $fin L^1(mathbb R)$ and $varphi$ a smooth compactly
supported function. Prove that $fast varphi$ is a smooth function
whose derivative is $fast varphi'$.




Any help would be much appreciated. I have been trying to work on this for the past couple days but all my ideas didn't lead me very far. Namely, I was trying to use the definition of a convolution. I am not sure though how the smooth and compactly supported $L^1$ function fits into this...






real-analysis functional-analysis hilbert-spaces convolution




real-analysis functional-analysis hilbert-spaces convolution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 at 18:09

























asked Dec 7 at 8:04









MathIsHard

1,215416




1,215416








  • 1




    For the first part, use the definition of the convolution (and maybe the identity $f ast g = g ast f$), and then differentiate under the integral sign (you need to show that this is allowed).
    – PhoemueX
    Dec 7 at 8:47










  • The direct approach is to find a bound for $f ast varphi' - f ast (varphi(.+y)-varphi)/y$. The less direct one is to look at $int f ast varphi'$
    – reuns
    Dec 7 at 16:44














  • 1




    For the first part, use the definition of the convolution (and maybe the identity $f ast g = g ast f$), and then differentiate under the integral sign (you need to show that this is allowed).
    – PhoemueX
    Dec 7 at 8:47










  • The direct approach is to find a bound for $f ast varphi' - f ast (varphi(.+y)-varphi)/y$. The less direct one is to look at $int f ast varphi'$
    – reuns
    Dec 7 at 16:44








1




1




For the first part, use the definition of the convolution (and maybe the identity $f ast g = g ast f$), and then differentiate under the integral sign (you need to show that this is allowed).
– PhoemueX
Dec 7 at 8:47




For the first part, use the definition of the convolution (and maybe the identity $f ast g = g ast f$), and then differentiate under the integral sign (you need to show that this is allowed).
– PhoemueX
Dec 7 at 8:47












The direct approach is to find a bound for $f ast varphi' - f ast (varphi(.+y)-varphi)/y$. The less direct one is to look at $int f ast varphi'$
– reuns
Dec 7 at 16:44




The direct approach is to find a bound for $f ast varphi' - f ast (varphi(.+y)-varphi)/y$. The less direct one is to look at $int f ast varphi'$
– reuns
Dec 7 at 16:44










1 Answer
1






active

oldest

votes


















1














I have an answer for the first question: recall:
$$f ast phi= phi ast f=int_mathbb{R}{phi(x-t)f(t)dt}$$
Note that in our case the convolution is defined everywhere since $phi$ is compactly supported and thus bounded, and $f in L^1$.
Now look at the definition of the derivative:
$$frac{f ast phi (x+h) - f ast phi (x)}{h} = int_mathbb{R}{frac{phi(x-t+h) - phi(x-t)}{h}f(t)dt}$$
By the mean value theorem for each fixed $t$ there exists $theta in [x-t+h,x-t]$ with $phi'(theta)=frac{phi(x-t+h) - phi(x-t)}{h}$. But since it is compactly supported we know that all derevatives of $phi$ are bounded by some constant $M$, thus we have that the absolute value of the integrand on the right is bounded (without dependence on h) by:
$$vert f(t)phi(theta)vertle Mvert f(t)vert$$
which is an integrable function. Now apply the Dominated convergence theorem on some sequence $h_nrightarrow 0$ and get:
$$underset{nrightarrow infty}{lim}frac{f ast phi (x+h_n) - f ast phi (x)}{h_n}=int_mathbb{R}{underset{nrightarrow infty}{lim}frac{phi(x-t+h_n) - phi(x-t)}{h_n}f(t)dt}$$
$$ = int_mathbb{R}{phi'(x-t) ast f(t)dt} = phi' ast f (x)$$.



Great, we found the derevative of $phi ast f$. Now since easily $phi'$ is again a compactly supported $C_infty(mathbb{R})$ function, we can apply the above argument again and again to get that $f ast phi$ is in $C_infty(mathbb{R})$.






share|cite|improve this answer























  • Thank you very much. I really appreciate the help on this.
    – MathIsHard
    Dec 9 at 6:35






  • 1




    Thanks very much for accepting my first answer on this site!
    – pitariver
    Dec 9 at 7:10











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1 Answer
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1 Answer
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active

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1














I have an answer for the first question: recall:
$$f ast phi= phi ast f=int_mathbb{R}{phi(x-t)f(t)dt}$$
Note that in our case the convolution is defined everywhere since $phi$ is compactly supported and thus bounded, and $f in L^1$.
Now look at the definition of the derivative:
$$frac{f ast phi (x+h) - f ast phi (x)}{h} = int_mathbb{R}{frac{phi(x-t+h) - phi(x-t)}{h}f(t)dt}$$
By the mean value theorem for each fixed $t$ there exists $theta in [x-t+h,x-t]$ with $phi'(theta)=frac{phi(x-t+h) - phi(x-t)}{h}$. But since it is compactly supported we know that all derevatives of $phi$ are bounded by some constant $M$, thus we have that the absolute value of the integrand on the right is bounded (without dependence on h) by:
$$vert f(t)phi(theta)vertle Mvert f(t)vert$$
which is an integrable function. Now apply the Dominated convergence theorem on some sequence $h_nrightarrow 0$ and get:
$$underset{nrightarrow infty}{lim}frac{f ast phi (x+h_n) - f ast phi (x)}{h_n}=int_mathbb{R}{underset{nrightarrow infty}{lim}frac{phi(x-t+h_n) - phi(x-t)}{h_n}f(t)dt}$$
$$ = int_mathbb{R}{phi'(x-t) ast f(t)dt} = phi' ast f (x)$$.



Great, we found the derevative of $phi ast f$. Now since easily $phi'$ is again a compactly supported $C_infty(mathbb{R})$ function, we can apply the above argument again and again to get that $f ast phi$ is in $C_infty(mathbb{R})$.






share|cite|improve this answer























  • Thank you very much. I really appreciate the help on this.
    – MathIsHard
    Dec 9 at 6:35






  • 1




    Thanks very much for accepting my first answer on this site!
    – pitariver
    Dec 9 at 7:10
















1














I have an answer for the first question: recall:
$$f ast phi= phi ast f=int_mathbb{R}{phi(x-t)f(t)dt}$$
Note that in our case the convolution is defined everywhere since $phi$ is compactly supported and thus bounded, and $f in L^1$.
Now look at the definition of the derivative:
$$frac{f ast phi (x+h) - f ast phi (x)}{h} = int_mathbb{R}{frac{phi(x-t+h) - phi(x-t)}{h}f(t)dt}$$
By the mean value theorem for each fixed $t$ there exists $theta in [x-t+h,x-t]$ with $phi'(theta)=frac{phi(x-t+h) - phi(x-t)}{h}$. But since it is compactly supported we know that all derevatives of $phi$ are bounded by some constant $M$, thus we have that the absolute value of the integrand on the right is bounded (without dependence on h) by:
$$vert f(t)phi(theta)vertle Mvert f(t)vert$$
which is an integrable function. Now apply the Dominated convergence theorem on some sequence $h_nrightarrow 0$ and get:
$$underset{nrightarrow infty}{lim}frac{f ast phi (x+h_n) - f ast phi (x)}{h_n}=int_mathbb{R}{underset{nrightarrow infty}{lim}frac{phi(x-t+h_n) - phi(x-t)}{h_n}f(t)dt}$$
$$ = int_mathbb{R}{phi'(x-t) ast f(t)dt} = phi' ast f (x)$$.



Great, we found the derevative of $phi ast f$. Now since easily $phi'$ is again a compactly supported $C_infty(mathbb{R})$ function, we can apply the above argument again and again to get that $f ast phi$ is in $C_infty(mathbb{R})$.






share|cite|improve this answer























  • Thank you very much. I really appreciate the help on this.
    – MathIsHard
    Dec 9 at 6:35






  • 1




    Thanks very much for accepting my first answer on this site!
    – pitariver
    Dec 9 at 7:10














1












1








1






I have an answer for the first question: recall:
$$f ast phi= phi ast f=int_mathbb{R}{phi(x-t)f(t)dt}$$
Note that in our case the convolution is defined everywhere since $phi$ is compactly supported and thus bounded, and $f in L^1$.
Now look at the definition of the derivative:
$$frac{f ast phi (x+h) - f ast phi (x)}{h} = int_mathbb{R}{frac{phi(x-t+h) - phi(x-t)}{h}f(t)dt}$$
By the mean value theorem for each fixed $t$ there exists $theta in [x-t+h,x-t]$ with $phi'(theta)=frac{phi(x-t+h) - phi(x-t)}{h}$. But since it is compactly supported we know that all derevatives of $phi$ are bounded by some constant $M$, thus we have that the absolute value of the integrand on the right is bounded (without dependence on h) by:
$$vert f(t)phi(theta)vertle Mvert f(t)vert$$
which is an integrable function. Now apply the Dominated convergence theorem on some sequence $h_nrightarrow 0$ and get:
$$underset{nrightarrow infty}{lim}frac{f ast phi (x+h_n) - f ast phi (x)}{h_n}=int_mathbb{R}{underset{nrightarrow infty}{lim}frac{phi(x-t+h_n) - phi(x-t)}{h_n}f(t)dt}$$
$$ = int_mathbb{R}{phi'(x-t) ast f(t)dt} = phi' ast f (x)$$.



Great, we found the derevative of $phi ast f$. Now since easily $phi'$ is again a compactly supported $C_infty(mathbb{R})$ function, we can apply the above argument again and again to get that $f ast phi$ is in $C_infty(mathbb{R})$.






share|cite|improve this answer














I have an answer for the first question: recall:
$$f ast phi= phi ast f=int_mathbb{R}{phi(x-t)f(t)dt}$$
Note that in our case the convolution is defined everywhere since $phi$ is compactly supported and thus bounded, and $f in L^1$.
Now look at the definition of the derivative:
$$frac{f ast phi (x+h) - f ast phi (x)}{h} = int_mathbb{R}{frac{phi(x-t+h) - phi(x-t)}{h}f(t)dt}$$
By the mean value theorem for each fixed $t$ there exists $theta in [x-t+h,x-t]$ with $phi'(theta)=frac{phi(x-t+h) - phi(x-t)}{h}$. But since it is compactly supported we know that all derevatives of $phi$ are bounded by some constant $M$, thus we have that the absolute value of the integrand on the right is bounded (without dependence on h) by:
$$vert f(t)phi(theta)vertle Mvert f(t)vert$$
which is an integrable function. Now apply the Dominated convergence theorem on some sequence $h_nrightarrow 0$ and get:
$$underset{nrightarrow infty}{lim}frac{f ast phi (x+h_n) - f ast phi (x)}{h_n}=int_mathbb{R}{underset{nrightarrow infty}{lim}frac{phi(x-t+h_n) - phi(x-t)}{h_n}f(t)dt}$$
$$ = int_mathbb{R}{phi'(x-t) ast f(t)dt} = phi' ast f (x)$$.



Great, we found the derevative of $phi ast f$. Now since easily $phi'$ is again a compactly supported $C_infty(mathbb{R})$ function, we can apply the above argument again and again to get that $f ast phi$ is in $C_infty(mathbb{R})$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 8 at 8:14

























answered Dec 8 at 8:07









pitariver

715




715












  • Thank you very much. I really appreciate the help on this.
    – MathIsHard
    Dec 9 at 6:35






  • 1




    Thanks very much for accepting my first answer on this site!
    – pitariver
    Dec 9 at 7:10


















  • Thank you very much. I really appreciate the help on this.
    – MathIsHard
    Dec 9 at 6:35






  • 1




    Thanks very much for accepting my first answer on this site!
    – pitariver
    Dec 9 at 7:10
















Thank you very much. I really appreciate the help on this.
– MathIsHard
Dec 9 at 6:35




Thank you very much. I really appreciate the help on this.
– MathIsHard
Dec 9 at 6:35




1




1




Thanks very much for accepting my first answer on this site!
– pitariver
Dec 9 at 7:10




Thanks very much for accepting my first answer on this site!
– pitariver
Dec 9 at 7:10


















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