Calculate convolution (integral) 
We will define the convolution (slightly unconventionally to match Rudin's proof) of $f$ and $g$ as follows:
  $$(f\star g)(x)=\int_{-1}^1f(x+t)g(t)\,dt\qquad(0\le x\le1)$$
  
  
*
  
*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!
 A: $f\star\delta_{10}(x) = \int_{-1}^1 f(x - t)\delta(t)\ dt$
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.
A: 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
$$
\left(f*\delta_{10}\right)(x) = \begin{cases}
5(\max\{x-1/10,\, 0.4\} - \min\{x+1/10,\, 0.7\}) &\text{if } x\in(0.3, 0.8) \\
0 &\text{otherwise}
\end{cases}
$$
A: 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 $n$.
