Suppose $g \in C^{1}[0,1]$ such that $$\int_{0}^{1}|g'(t)|^2\:dt \leq R$$ for some constant $R$.
I am interested in proving that for all $a,b \in [0,1]$, $$|g(a)-g(b)|\leq \sqrt{R}|a-b|^{1/2}.$$
I tried using the Mean Value Theorem to show that $$|g(a)-g(b)|\leq \sqrt{R}|a-b|$$ but it's missing $|a-b|^{1/2}$.
Any hint or help will be appreciated. Thanks.
Define $f(t)=\dfrac{g(t)}{\sqrt{R}}$. Then we have by the assumption that
$\int_{0}^{1}|f'(t)|^{2}dt\leq\,1$. Take $a<b$ in $[0,1]$. Then
$f(b)-f(a)=\int_{a}^{b}f'(t)dt$ and hence
$|f(b)-f(a)|\leq\,\int_{a}^{b}|f'(t)|dt$. Now we use the Cauchy-Schwartz inequality for $F(t)=|f'(t)|$ and $G(t)=1$ and we get:
$\int_{a}^{b}|f'(t)|dt\,\leq\,(\int_{a}^{b}|f'(t)|^{2}dt)^{1/2}(b-a)^{1/2}$$\leq\,\int_{0}^{1}|f'(t)|^{2}dt$$(b-a)^{1/2}$$\leq\,(b-a)^{1/2}$.
Therefore we get $|\dfrac{g(b)}{\sqrt{R}}-\dfrac{g(a)}{\sqrt{R}}|\,\leq\,(b-a)^{1/2}$
which implies $|g(b)-g(a)|\leq\,\sqrt{R}(b-a)^{1/2}$