Calculate convolution (integral)

Question

We will define the convolution (slightly unconventionally to match Rudin's proof) of $f$ and $g$ as follows: $$(f\star g)(x)=\int_{-1}^1f(x+t)g(t)\,dt\qquad(0\le x\le1)$$

1. Let $\delta_n(x)$ be defined as $\frac n2$ for $-\frac1n<x<\frac1n$ and 0 for all other $x$. Let $f(x)$ be defined as $x$ for $.4<x<.7$ and let $f(x)=0$ for all other $x$. Find a piecewise algebraic expression for $f\star\delta_{10}$ and graph $f\star\delta_{10}$. Repeat the exercise for $f\star\delta_{20}$. In what sense does $f\star\delta_n$ converge on $[0,1]$ and to what function does it converge?

Hello everyone, I just need help finding a piecewise algebraic expression for $f* \delta_{ 10}$. I think I should be able to figure out everything else in the question once I know how to do this.

Thoughts/things I know (how to do): Delta 10 is defined as $5$ for $-1/10<x<1/10$ and $0$ otherwise. $f(x)=x$ for $0.4<x<0.7$ and $0$ otherwise. I know how both graphs look like visually. I guess you could say there is a jump in the graph of f(x) at 0.4 and 0.7 and a jump in the graph of $\delta_{10}$ at $-1/10$ and $1/10$.

Now confusions: I believe that perhaps I will have to split up the problem into several cases/integrals as both f(x) and delta are piecewise. The main problem in this question is I don't understand what the t represents so I don't know how to set up my bounds as my bounds are in terms of t (as we integrate with respects to $dt$). I believe that the integral seemingly from the definition is only valid for $0\leq x<\leq1$ inclusive. Also if I'm doing $f* \delta_{10}$, lets say for x between $0.4<x<0.7$ then wouldn't $f(x+t)=x+t$ and $g(t)=0$? Overall, I think I'm confused so I could really use some guidance on this problem for the first case $f*\delta_{10}$, then I believe I could figure out the rest. Thank you!

Answer 1

$f\star\delta_{10}(x) = \int_{-1}^1 f(x - t)\delta(t)\ dt$

but $\delta_{10}(t) = 0$ for most of this interval

$f\star\delta_{10}(x) = \int_{-\frac {1}{10}}^{\frac {1}{10}} 5(x - t) dt$

$x\in [0.5, 0.6]$

But if $x$ is outside this interval:

$x\in [0.3, 0.5]$

$f\star \delta_{10}\int_{0.4-x}^{0.1} 5(x - t) dt$

$x\in [0.6, 0.8]$

$f\star \delta_{10}\int_{-0.1}^{0.7-x} 5(x - t) dt$

$f\star\delta_{10} = \begin {cases} \frac 52 (x+0.1)^2 - 0.4 &x\in[0.3,0.5]\\x&x\in[0.5,0.6]\\ -\frac 52x^2 + 0.5x + 1.2& x\in [0.6,0.8]\\0&\text {elsewhere}\end{cases}$

The delta function puts a small blur on the edges that sharpens as $n$ gets big.

Answer 2

By definition of convolution and $\delta_{10}$, we have; \begin{align*} \left(f*\delta_{10}\right)(x) = \int_{-1}^1 f(x+t)\delta_{10}(t)\, dt =5\int_{-1/10}^{1/10} f(x+t)\, dt =5\int_{x-1/10}^{x+1/10} f(t)\, dt \end{align*} Now, $f(t) = 0$ outside of the interval $(0.4, 0.7)$. Thus, the above integral evaluates to $0$ is either $x-1/10 \geq 0.7 \iff x\geq 0.8$ or $x+1/10 \leq 0.4 \iff x\leq 0.3$. We are therefore only concerned with the case when $x\in (0.3, 0.8)$. Assuming that $x$ is in that interval, we have \begin{align*} \int_{x-1/10}^{x+1/10} f(t)\, dt &=\int_{\max\{x-1/10,\, 0.4\}}^{\min\{x+1/10,\, 0.7\}} f(t)\, dt =\int_{\max\{x-1/10,\, 0.4\}}^{\min\{x+1/10,\, 0.7\}} 1\, dt\\ &=\max\{x-1/10,\, 0.4\} - \min\{x+1/10,\, 0.7\} \end{align*} We conclude that $$ \left(f*\delta_{10}\right)(x) = \begin{cases} 5(\max\{x-1/10,\, 0.4\} - \min\{x+1/10,\, 0.7\}) &\text{if } x\in(0.3, 0.8) \\ 0 &\text{otherwise} \end{cases} $$

Answer 3

By definition, $$ f*\delta_{10}=\int_{-1}^1f(x+t)\delta_{10}(t)\mathrm dt\\ =5\int_{-.1}^{.1}f(x+t)\mathrm dt $$ Now we use the argument of $f$ and the definition of $f$ to determine the rest.

The integrand vanishes (by the definition of $f$) unless $.4<x+t<.7$, i.e. $$ .4-x<t<.7-x $$ Noting that the size of this interval, $.7-x-.4+x=.3$, is larger than the length of the interval we are integrating over, $.2$.

We shift the window for $t$ around to figure out what the integral is in each case.

If the window misses the entire interval, i.e. if $.7-x\leq -.1$ or if $.4-x\geq .1$., the above integral vanishes. If instead the right hand side of the interval intersects the domain of integration, we have $$ -.1\leq .7-x\leq .1 $$ then the convolution is the integral $$ 5\int_{-.1}^{.7-x}(x+t)\mathrm dt $$ and if the left hand side does, $-.1\leq .4-x\leq .1$, then the convolution is $$ 5\int_{.4-x}^{.1}(x+t)\mathrm dt $$ if we the left and right end points straddle the entire interval, i.e. $.4-x\leq -.1$ and $.7-x\geq .1$, then we have $$ 5\int_{-.1}^.1(x+t)\mathrm dt=x $$ Putting this all together in piecewise function form we have $$ f*\delta_{10}(x)= \begin{cases} 0&x\leq .3\\ 5\int_{.4-x}^{.1}(x+t)\mathrm dt&.3\leq x\leq .5\\ x&.5\leq x\leq .6\\ 5\int_{-.1}^{.7-x}(x+t)\mathrm dt&.6\leq x\leq .8\\ 0&x\geq .8 \end{cases} $$ Where I leave to you to simplify the integrals.

As for the latter part of the question, a hint may be that $$ \lim_{n\to \infty}=\delta_n=\begin{cases}\infty&x=0\\ 0&x\ne 0\end{cases} $$ and $\int_{\mathbb{R}}\delta_n\mathrm dx=1$ for any $n$.