-3
$\begingroup$

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,

$\endgroup$
1
  • 3
    $\begingroup$ $H^1$ functions are almost everywhere differentiable, so any nowhere differentiable continuous function with compact support is a counterexample. $\endgroup$ Feb 17, 2014 at 13:07

1 Answer 1

3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.