Calculation of convolution of two indicator functions So the problem I'm working on states...
Let $f_k=\mathbf{1}_{[-1,1]}\ast\mathbf{1}_{[-k,k]}$, where $\ast$ is a convolution. Compute $f_k(x)$ explicitly.
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I know that the value of the function is dependent on x. However, I've been working on this for the better part of the day and can't come up with anything reasonable.
I know that the function is nonzero only when $x-1\leq y\leq x+1$ and $-k\leq y\leq k$.
I have that (only because I saw something similiar here):
\begin{eqnarray*}
f_k(x) &=& \int_\mathbb{R}\mathbf{1}_{[-1,1]}(x-y)\cdot\mathbf{1}_{[-k,k]}(y)\ dy\\
&=& \int_{-k}^k\mathbf{1}_{[-1,1]}(x-y)\ dy = \int_{-k}^k\mathbf{1}_{[-1,1]}(y-x)\ dy \\
&=& \int_{-k}^k\mathbf{1}_{[x-1,x+1]}(y)\ dy
\end{eqnarray*}
Will this work? I still don't know how to break $\mathbb{R}$ into whatever intervals of $x$. Could someone give me any hints or suggestions on how to proceed with this problem?
 A: I'm continuing your evaluation of the integral: 
$$=\int_{\mathbb{R}}1_{[-k,k]}(y)\cdot1_{[x-1,x+1]}(y)dy=\int_\mathbb{R}1_{[-k,k]\cap[x-1,x+1]}(y)dy=m([-k,k]\cap[x-1,x+1])$$
where $m$ denotes the Lebesgue measure. Now all you have to do is determine what does that intersection look like, so you'll have to distinguish cases:
Case 1:  $x\in(-\infty,-k-1]$. Then $x-1\leq x+1\leq -k$, therefore the intersection is at most $\{k\}$ which has measure $0$.
Case 2: $x\in(-k-1,-k+1]$. Then $x-1\leq-k\leq x+1\leq k$, therefore the intersection is the segment $[-k,x+1]$ which has measure $x+k+1$.
Case 3: (Here I assume $k\geq 1$. The case $k\in(0,1)$ is treated later on) $x\in(-k+1,k-1]$. Then $-k\leq x-1\leq x+1\leq k$ therefore the intersections is $[x-1,x+1]$, which has measure $2$.
Case 4: $x\in[k-1,k+1]$. Then $-k\leq x-1\leq k\leq x+1]$ therefore the intersection is $[x-1,k]$ which has measure $k+1-x$.
Case 5: $x\in (k+1,\infty)$: Then $-k<k<x-1<x+1$ so the intersection is empty, that has measure $0$.
The convolution is the piecewise function defined on the intervals above as the measures that were computed.
Now if $k\in(0,1)$, Case 3 does not appear. That is, if $x\in(-k+1,k+1]$, then $k<x+1$. Indeed, if $x+1<k$, then $x<k-1<0$, but $x>1-k>0$ a contradiction. Therefore it is $-k<x-1<k<x+1$ and the intersection is $[x-1,k]$ with measure $k+1-x$.
Conclusion: this convolution is $0$ from $-\infty$ to a certain point. In a following interval it start increasing linearly with factor $1$ until a certain point. Then, if $k\geq1$, it remains constant on an interval and then starts decreasing linearly with factor $-1$. If $k<1$, it does not remain constant and immediately starts decreasing linearly with factor ${-1}$. When it hits $0$, in both cases remains constantly $0$ on the rest of the real line.
