Let $(f\ast g)(x)=\int f(y)g(x-y)dy$, $\chi_1=\chi_{[-1,1]}$ and $\chi_n=\chi_{[-n,n]}$ for $n\in \mathbb N$.
I am trying to calculate $f_n=\chi_1\ast \chi_n$ and show $\|f_n\|_ \infty=2$. Since $\chi_n(x-y)$ is $1$ on $[x-n,x+n]$ and $0$ otherwise, I calculated
$$ \begin{aligned}f_n(x)&=(\chi_1\ast \chi_n)(x)= \int\chi_{[-1,1]\cap[x-n,x+n]}(y)\\[0.2cm] &=\begin{cases}0 & \text{if} \quad x<-1-n\\ 1+x+n & \text{if} \ -1-n\leq x\leq0 \\ 1+n-x& \text{if} \quad 0\leq x\leq 1+n \\ 0 & x>n+1 \\ \end{cases} \end{aligned}$$
However in this case how can $\|f_n\|_ \infty=2$ be possible? I get supremum as $1+n$.
I have some mistakes but I am not able to see that.
I am sorry for this easy question and thanks a lot for any help.
\begin{equation*} \begin{split} (\chi_1*\chi_n)(x) & = \int_\mathbb{R}\chi_{[-1,1]}(y)\chi_{[-n,n]}(x-y)dy\\ & = \int_{-1}^1 \chi_{[-n,n]}(x-y)dy \\ & = \int_{x+1}^{x-1}\chi_{[-n,n]}(t)(-dt) \\ & =\int_{x-1}^{x+1}\chi_{[-n,n]}(t)dt \\ & = \mu([x-1,x+1]\cap[-n,n]), \end{split} \end{equation*}
where $\mu$ is the usual Lebesgue measure (i.e. the length of the intersection) and where we used the substitution $t=x-y$ in the third line. Clearly the length cannot be bigger than $2$ (since $\mu([x-1,x+1])=2$), and the value $2$ is attained for every $n$ when $x=0$, since $[-1,1]\subset[-n,n]$ for any natural $n$. Therefore $\|\chi_1*\chi_n\|_\infty=2$ for all $n$.