Let $f$ be a continuous function on $[0,\infty)$ and let $C\geq 0$ be a constant such as $$ \int_x^{x+1} |f(t)|\,\mathrm{d} t\leq C\ \ ,\ \forall x\geq 0. $$ Prove that
$$ e^{-x}\int_0^x e^t|f(t)|\,\mathrm{d} t\leq C\cdot\frac{e}{e-1}\ \ ,\ \forall x\geq 0. $$
PROOF
So far I have tried to seperate the first integral in two integrals and use integration by parts. My next thought was to use absolute value inequalities. I think this method is very wrong. Any help?
Make the substitution $u = x-t$. Your integral expression becomes
$$\int_0^x e^{-u} \lvert f(x-u)\rvert\,du.$$
For $k < \lfloor x\rfloor$, estimate
$$\int_k^{k+1} e^{-u}\lvert f(x-u)\rvert\,du \leqslant e^{-k} \int_k^{k+1} \lvert f(x-u)\rvert\,du \leqslant C e^{-k},$$
and for $k = \lfloor x\rfloor$, estimate
$$\int_k^x e^{-u}\lvert f(x-u)\rvert\,du \leqslant e^{-k} \int_k^x \lvert f(x-u)\rvert\,du \leqslant C e^{-k}.$$
Summing up, we obtain
$$\int_0^x e^{-u} \lvert f(x-u)\rvert\,du \leqslant C\cdot \sum_{k = 0}^{\infty} e^{-k} = C\frac{e}{e-1}.$$