# $L^{1}(\mathbb R) \cap L^{2}(\mathbb R) \cap C_{0}(\mathbb R)\subset H_{1}(\mathbb R)$?

Put, $C_{0} (\mathbb R)=\{f:\mathbb R \to \mathbb C: f \text { is continuous on} \ \mathbb R \ \text {and } \lim_{|x|\to \pm \infty}f(x)=0 \}$(= Continuous functions on $\mathbb R$ vanishing at $\infty$) and $A:= L^{1}(\mathbb R) \cap L^{2}(\mathbb R) \cap C_{0}(\mathbb R).$

My Question: Given $|f|\in A$. Can we always expect $|f|\in H_{1}(\mathbb R)=\{ f\in \mathcal{S'}(\mathbb R): (1+|\xi|^{2})^{1/2}\hat{f}\in L^{2}(\mathbb R) \}$ (Sobolev spaces), Or, we we can produce a counter example ?

Thanks,

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$H^1$ functions are almost everywhere differentiable, so any nowhere differentiable continuous function with compact support is a counterexample. – Alexander Shamov Feb 17 '14 at 13:07

No. A simple family of counter examples come from the Sobolev embeddings. Since $$W^{1,2}(\mathbb{R})\hookrightarrow C^{0,\frac{1}{2}}(\mathbb{R}),$$ your identity would imply that continuous functions are always Hölder continuous.