Let $H^{1}(a, b)=W^{1,2}(a,b)$ be the Sobolev space, that is, the functions $f\in L^2$ such that $f' \in L^2$, where $f'$ denotes the weak derivative of $f$. This is equivalent to $$f(x)=c+\int_a^x f'(x) dx $$ for some $c \in \mathbb{R}$, almost everywhere in $(a,b)$. Prove that if $f \in H^{1}(a, b)$, then \begin{equation} |f(t)-f(s)| \leq\left\|f^{\prime}\right\|_{L^{2}(a, b)}|t-s|^{1 / 2}, \end{equation} that is, $f$ is Hölder continuous. Also show that if $s \in (a, b)$, then $$ \lim _{t \rightarrow s} \frac{|f(t)-f(s)|}{|t-s|^{1 / 2}}=0 . $$
I managed to prove the first part, but for the second I'm stuck. I tried using that $$\lvert f(t)-f(s)\rvert=\left\lvert \int_s^t f'(x) dx\right\rvert \leq \int_s^t \lvert f'(x) \rvert dx \leq (t-s)\operatorname{sup}_{x \in [a,b]}|f'(x)|,$$ but $f' \in L^2$ is not necessarily bounded.
In the last step, you can use Cauchy-Schwarz inequality $$ \int_s^t |f'(x)|\,dx\le \left(\int_s^t |f'(x)|^2\,dx\right)^{1/2}\cdot\left(\int_s^t 1^2\,dx\right)^{1/2}=\left(\int_s^t |f'(x)|^2\,dx\right)^{1/2}\cdot|t-s|^{1/2}. $$ As $s\to t$ the integral on the right-hand side approaches zero, by (absolute) continuity of the Lebesgue integral.