Let $g \in L^p([a,b])$ for some $p \in (1, \infty]$. Define $$f(x) = f(a) + \int_a^xg(s)\ ds$$ for any $x \in [a,b]$. The claim is that $f$ is $\frac{1}{p^{\prime}}$-Hölder continuous on $[a,b]$, where $\frac{1}{p} + \frac{1}{p^{\prime}} = 1$.
So I need to show that there exists $M > 0$ and $\delta > 0$, such that for all $t$ with $|t|< \delta$, $|f(x+t) - f(x)| \leq M|t|^{1/p^{\prime}}.$ We have $$|f(x+t) - f(x)| = \left|\int_a^{x+t}g(s)\ ds - \int_a^xg(s)\ ds\right| \leq \int_a^{x+t}|g(s)|\ ds + \int_a^x|g(s)|\ ds. $$ What should I do next? I seem to cannot find a way to use the fact that $g \in L^P([a,b])$. I must say that the condition $$\frac{1}{p} + \frac{1}{p^{\prime}} = 1$$ is very similar to Hölder's inequality. Should I use that too somewhere in the proof?
EDIT: So after reading the comments, I decided to add some new discoveries. We have $$|f(x+t) - f(x)| = \left|\int_x^{x+t}g(s)ds\right|\leq \int_x^{x+t}|g(s)|ds.$$ Rewriting $\int_x^{x+t}|g(s)|ds$ into $$\left(\int_x^{x+t}|g(s)|^{p\cdot\frac{1}{p}} \right)^{p\cdot\frac{1}{p}} = || g^{1/p}||_p^p = ||1 \cdot g^{1/p}||_p^p \leq ||g^{1/p}||_p^p||1||_{p^{\prime}}^p.$$ Still, I don't see that this proves $\frac{1}{p^{\prime}}$-Hölder continuous$. I don't think the rewriting part is correct.
Without loss of generality, let $t>0$: As Teresa said: $$ \lvert f(x+t) - f(x) \rvert \leq \int^{x+t}_x \lvert g(s) \rvert~\mathrm{d}s = \lVert 1 \cdot g \rVert_{L^1[x, x+t]} $$ Now use Hölder's inequality: (Let $M := \lVert g \rVert_{L^{p}[a, b]}$) $$ \lVert 1 \cdot g \rVert_{L^1[x, x+t]} \leq \lVert 1 \rVert_{L^{p'}[x, x+t]} \lVert g \rVert_{L^{p}[x, x+t]} = \lVert g \rVert_{L^{p}[x, x+t]} \underbrace{\lvert t \rvert}_{=t}~^{\frac{1}{p'}} \leq M \lvert t \rvert^{\frac{1}{p'}} $$