Let $ \varphi$ a bump function with support $[-\frac {1}{2},\frac {1}{2}] $. and
$ \mathbf{1}_{[-\frac {1}{2},\frac {1}{2}]} (x) =\begin{cases} 1 & \text{if }x \in [-\frac {1}{2},\frac {1}{2}] \\ 0 & \text{if }x \notin [-\frac {1}{2},\frac {1}{2}]. \end{cases}$
Let $f=\mathbf{1}_ {[-\frac {1}{2},\frac {1}{2}]}* \varphi $ (* convolution ).
Then Support of $f$ is $[-1, 1] $
Show that
$\frac{1}{f(0)} \displaystyle\sum_{n=-\infty}^{+\infty} f(x-n)=1$
Any hint. Maybe I use Dirac comb
Fix a point $x$ and look closely at $f(x-n)$: \begin{align*} f(x-n) &= \int_{-\infty}^\infty \boldsymbol{1}_{[-1/2,1/2]}(x-n-t) \phi(t) \, dt \\ &= \int_{-\infty}^\infty \boldsymbol 1_{[x-n-1/2,x-n+1/2]}(t) \phi(t) \, dt \\ &= \int_{-\infty}^{\infty} \boldsymbol 1_{[x-n-1/2,x-n+1/2]}(t) \boldsymbol 1_{[-1/2,1/2]}(t) \phi(t) \, dt. \end{align*}
The intervals $\{[x-n-1/2,x-n+1/2]\}$ each have length $1$ and are thus nonoverlapping. Consequently $$\sum_{n = -\infty}^\infty \boldsymbol 1_{[x-n-1/2,x-n+1/2]}(t) = 1$$ for almost all $t \in \mathbb R$. Apply an appropriate convergence theorem to determine $$ \sum_{n=-\infty}^\infty f(x-n) = \int_{-\infty}^\infty \boldsymbol 1_{[-1/2,1/2]}(t) \phi(t) \, dt$$ where $$ \int_{-\infty}^\infty \boldsymbol 1_{[-1/2,1/2]}(t) \phi(t) \, dt = \int_{-\infty}^\infty \boldsymbol 1_{[-1/2,1/2]}(-t) \phi(t) \, dt = f(0). $$