I am working on this
Suppose that $f,f'\in L^1(\mathbb{R})$. Then $\sum_{k= 0} ^{\infty}\vert f(x+k)\vert\in L^{\infty}([a,b])$ for any $a,b\in \mathbb{R}.$
Idea: Let $i$ be any integer. $\int_i ^{i+1}\vert \sum_k f(x+k) \vert dx=\sum_k \int _i ^{i+1}\vert f(x+k)\vert=\sum_k \int _{i+k} ^{i+1+k}\int \vert f(x) \vert \\\le \int_{-\infty} ^{\infty}\vert f\vert <\infty$ since $f$ is in $L^1$.
How do I use the fact that $f'$ is integrable, and how can I show the series belongs to $L^{\infty}([a,b])$?
We have to approximate $f(x)$ by integrals involving $f$ and $f'$. We actually have $$\left|f(x)-\int_x^{x+1}f(t)\mathrm dt \right|\leqslant \int_x^{x+1}\left| f'(t)\right| \mathrm dt. $$ This follows from the fundamental theorem of calculus. Indeed, $$\int_x^{x+1}f(t)\mathrm dt-f(x) =\int_0^1 (f(x+t)-f(x))\mathrm dt=\int_0^1\int_0^t f'(x+s)\mathrm ds\mathrm dt,$$ then switch the integrals.
As a consequence, for each $x\in\mathbf R$, $$|f(x)|\leqslant \int_x^{x+1}\left| f(t)\right| \mathrm dt+\int_x^{x+1}\left| f'(t)\right| \mathrm dt,$$ hence $$\sum_{j=0}^{\infty}|f(x+j)|\leqslant\int_{x}^{\infty} \left| f(t)\right| \mathrm dt+ \int_{x}^{\infty}\left| f'(t)\right| \mathrm dt .$$