# Toronto Math Forum

## APM346-2012 => APM346 Math => Term Test 2 => Topic started by: Victor Ivrii on November 15, 2012, 08:14:38 PM

Title: TT2--Problem 1
Post by: Victor Ivrii on November 15, 2012, 08:14:38 PM
Let $f:{\mathbb{R}}\rightarrow {\mathbb{R}}$ be a continuous but non-differentiable function which satisfies $f(x)=0$ for all $|x| > 1.$ Let $g:{\mathbb{R}}\rightarrow {\mathbb{R}}$ be a continuous function which satisfies satisfies $g(x)=0$ for all $|x| > 2.$ Suppose further that derivative $g'$ and second derivative $g''$ are both continuous. The convolution $f*g$ of these two functions is defined by the formula
$$(f*g) (x) = \int f(x-y)g(y)\,dy.$$
• (a) Prove that the function $f*g(x) =0$ for $|x|>3$.
• (b) Prove that the derivative of the function $f*g$ is continuous.

post after 22:30
Title: Re: TT2--Problem 1
Post by: Jinchao Lin on November 15, 2012, 10:30:03 PM
Solution for part(a)
Title: Re: TT2--Problem 1
Post by: Ian Kivlichan on November 15, 2012, 10:31:55 PM
Hopeful solutions to both parts attached! :)
Title: Re: TT2--Problem 1
Post by: Chen Ge Qu on November 15, 2012, 10:32:44 PM
See attached
Title: Re: TT2--Problem 1
Post by: Ian Kivlichan on November 15, 2012, 10:38:38 PM
Chen Ge, in 1.b) I'm not sure you can have the derivative of f since it isn't differentiable.. :S
Title: Re: TT2--Problem 1
Post by: Victor Ivrii on November 16, 2012, 06:34:45 AM
(a) Actually it reflects the following property of convolution which we never mentioned:
$\newcommand{\supp}{\operatorname{supp}}$

\supp (f*g) \subset \supp(f)+\supp(g)
\label{eq-1}

where $\supp{f}$ is support of $f$ the smallest closed set outside of which $f=0$ and $+$ mean arithmetic sum of sets: $A+B= \{x+y :\,x\in A,\, y\in B\}$

(b) Ian is correct that one should not differentiate non-differentiable "factor" in convolution but instead first change variable of integration $y\mapsto x-y$

(f*g): = \int  f(x-y)g(y)\,dy = \int  f(y)g(x-y)\,dy
\label{eq-2}

and then differentiate:

(f*g)'=\left\{\begin{aligned}
&f'*g,
&f*g'
\end{aligned}\right.
\label{eq-3}

Note, I did not use $f*g=g*f$ as this would fail for (say) matrix valued functions while (\ref{eq-3}) would remain valid.

PS. Still, in the framework of the theory of distributions we can differentiate everything and

(f*g)'=f'*g=f*g'

where both differentiation and convolution are extended to this framework.

Title: Re: TT2--Problem 1
Post by: Heqian Zhang on November 16, 2012, 01:44:20 PM
Hi, professor. I did use Fourier Transform to prove the derivative of the convolution is continuous. Is that okï¼Ÿ
Title: Re: TT2--Problem 1
Post by: Victor Ivrii on November 16, 2012, 02:58:03 PM
Hi, professor. I did use Fourier Transform to prove the derivative of the convolution is continuous. Is that okï¼Ÿ

I don't know all the details of it, so I cannot say that it is OK. The correct solution via F.T. is possible, but even if correct it is not a solution I would like. The reason: to prove that $f*g$ is continuous you don't need to assume that both $f,g$ (and thus $f*g$) have good F.T. In fact everything makes perfect sense if $f$ is continuous, $g$ bounded and one of these functions is $0$ as $|x|\ge c$. In this case only this function necessarily has F.T., another function and their convolution may grow fast as $x\to \infty$ and not have F.T.

Title: Re: TT2--Problem 1
Post by: Thomas Nutz on December 16, 2012, 01:30:08 PM
Is eq. (3)
$$(f \star g)'=f\star g'$$ sufficient to prove that $(f \star g)'$ is continuous? I.e. can we assume that the convolution of two continuous functions is continuous, or do we have to take difference quotients as Ian did in his solution?
Title: Re: TT2--Problem 1
Post by: Victor Ivrii on December 16, 2012, 02:17:45 PM
Is eq. (3)
$$(f \star g)'=f\star g'$$ sufficient to prove that $(f \star g)'$ is continuous? I.e. can we assume that the convolution of two continuous functions is continuous, or do we have to take difference quotients as Ian did in his solution?

First, it is not a $\star$ but $*$. Second, you need to prove that the convolution of two continuous functions is continuous; actually, one function should be just integrable.