If $u\in C^{\beta}[0,a]$ for $0<\beta<1$ then prove that $\int_{0}^{x} u(t) dt \in C^{\beta+1}[0,a]$.
where $C^{\beta}[0,a]$ is space of Holder Continuous functions in [0,a].
$\textbf{I TRIED}$ $u\in C^{\beta}[0,a]$ hence $[u]_{\beta}=Sup \frac{\left|u(x)-u(y)\right|}{\left| x-y\right|^{\beta}}$ is finite for $0<\beta<1, x\neq y$
Consider $[\int_{0}^{x}u(t)dt]_{\beta}=Sup_{x,y\in[0,a]} \frac{\left|\int_{0}^{x}u(t)dt-\int_{0}^{y}u(t)dt\right|}{\left| x-y\right|^{\beta}}$
Overall I am not able to prove this, I tried a lot. Believe me I am trying hard to understand pure Mathematics. But even these small proofs also I am not able to do. I am surprised whether this is happening only with me or there are more people like me.