Let $f$ be a continuous, decreasing and integrable function on $\mathbb R^+$. Show the existence of a function $g$ continuous such that.
$g(x+1) - g(x)= f(x)$
All I can see is that :
$f$ is decreasing therefore admits a limit $l$ in $ +\infty$ in $\mathbb R \cup \{-\infty \}$. In addition, $f ∈ L^{1}(\mathbb R^+)$ So $l$ must be $0$.
I add a non-finished proof, it must be polished.
Since $f \in L^1(\mathbb R^+)$ and decreasing it follows that $f$ is nonnegative, $ \sum_{n=0}^\infty f(x+n) $ converges for every $x$, and also $\lim_{x\to+\infty} f(x)=0$. Then,
\begin{multline*} f(x)=f(x)-\lim_{t\to+\infty} f(t) = \sum_{n=0}^\infty (f(x+n)-f(x+(n+1)) \\= \sum_{n=0}^\infty f(x+n)-\sum_{n=0}^\infty f((x+1)+n) = g(x+1) - g(x), \end{multline*} where $$ g(x) = - \sum_{n=0}^\infty f(x+n). $$ We must prove that $g$ is continous, that is, we must prove that the series that defines $g$ converges uniformly. Indeed, since $f$ is decreasing, we have $$ \sum_{n=k}^l f(x+n) \leq \int_{k-1}^l f(t) dt \xrightarrow{l,k \to \infty} 0 \quad \forall x, $$ which proves that the series converges uniformly and therefore $g$ is continous.