Sobolev space $W^{1, 2}$, bounding $|f(x) - f(y)|$ by product of $L^2$ norm of derivative of $f$ and $|x - y|^{1/2}$.

Question

Let$$W^{1, 2}(\mathbb{R}) := \left\{ (f, f') \in L^2(\mathbb{R}) \times L^2(\mathbb{R}): f \text{ has a continuous representative and for all }x,\,y \in \mathbb{R},\text{ }f(x) - f(y) = \int_y^x f'(z)\,dz\right\}.$$If $(f, f') \in W^{1, 2}$, then for all $x$, $y \in \mathbb{R}$, do we have$$|f(x) - f(y)| \le \|f'\|_{L^2} |x - y|^{1/2}?$$

Answer 1

From your very definition of $W^{1,2} (\Bbb R)$ you have that

$$|f(x) - f(y)| = \left| \int \limits _y ^x f'(z) \ \Bbb d z \right| = | \langle f', 1 \rangle _{L^2} | \le \| f' \| _{L^2} \ \| 1 \| _{L^2} = \\ \| f' \| _{L^2} \sqrt {\int \limits _y ^x 1^2 \ \Bbb d x} = \| f' \| _{L^2} \sqrt{|x-y|} ,$$

where I have used the Cauchy-Schwarz inequality in $L^2(\Bbb R)$.