Morrey's Inequality in 1D

Question

Morrey's Inequality in 1D for $p=2$: There exists a constant $C$ such that

$\|u\|_{C^{0,1/2}(\mathbb{R})} \leqslant C \|u\|_{H^{1}(\mathbb{R})}$

for all $u \in C^{1}(\mathbb{R})$.

Of course, for any $u \in C^{1}$ we have by the fundamental theorem of calculus and Holder's inequality:

$|u(x)-u(y)| \leqslant \int_{y}^{x} |u'(t)|\;dt \leqslant \|u'\|_{L^{2}(\mathbb{R})} |x-y|^{1/2}$.

This implies $[u]_{C^{0,1/2}(\mathbb{R})} \leqslant \|u\|_{H^{1}(\mathbb{R})}$. To conclude, we need to show that $|u(x)| \leqslant C\|u\|_{H^{1}(\mathbb{R})}$ for all $x \in \mathbb{R}$. How can I go about proving this? I know Evans has a proof for $\mathbb{R}^{n}$ but I would like to know if there is a simpler approach in 1D.

Answer 1

Fix $x\in\mathbb{R}$. Then: $$\forall y\in\mathbb{R}, |u(x)|\le|u(y)|+|u(y)-u(x)|\le |u(y)|+\|u'\|_{L^2(\mathbb{R})}|x-y|^{1/2}.$$ Integrate this relation in $\operatorname{d}y$ for $y\in[x-1,x+1]$. Then, using Hölder inequality: $$2|u(x)|=\int_{[x-1,x+1]}|u(x)|\operatorname{d}y\le\int_{[x-1,x+1]}\left(|u(y)|+\|u'\|_{L^2(\mathbb{R})}|x-y|^{1/2}\right)\operatorname{d}y \\ \le \sqrt2\|u\|_{L^2([x-1,x+1])}+2\|u'\|_{L^2(\mathbb{R})} \le \sqrt2\|u\|_{L^2(\mathbb{R})}+2\|u'\|_{L^2(\mathbb{R})}\le2\|u\|_{H^1(\mathbb{R})}$$ So: $$|u(x)|\le\|u\|_{H^1(\mathbb{R})}$$ Being $x$ arbitrarily chosen, the conclusion follows.