0
$\begingroup$

I know that the following inequality is true for all $g\in C^{\infty}_{0}(\mathbb{R}^{3})$:

$$\biggl\Vert\frac{g}{\vert\cdot\vert}\biggr\Vert_{L^{2}(\mathbb{R}^{3})}\leq2\Vert \nabla g\Vert_{L^{2}(\mathbb{R}^{3})}$$

My goal is to show that this inequality is still true for some $f\in H^{1}(\mathbb{R}^{3})$.

My Ansatz so far: I use the fact that ${C^{\infty}_{0}(\mathbb{R}^{3})}$ is dense in $H^{1}(\mathbb{R}^{3})$, which means that we can find for each $\varepsilon\in\mathbb{R}_{>0}$ a function $g\in C_{0}^{\infty}(\mathbb{R}^{3})$, such that $$\Vert f-g\Vert_{H^{1}(\mathbb{R}^{3})}=\sqrt{\Vert f-g\Vert_{L^{2}(\mathbb{R}^{3})}^{2}+\sum_{i=1}^{3}\Vert \partial_{i}(f-g)\Vert_{L^{2}(\mathbb{R}^{3})}^{2}}\leq\varepsilon.$$ This means that we must have that

$$\Vert f-g\Vert_{L^{2}(\mathbb{R}^{3})}\leq \varepsilon\hspace{1cm}\mathrm{and}\hspace{1cm}\Vert \nabla (f-g)\Vert_{L^{2}(\mathbb{R}^{3})}\leq \sum_{i=1}^{3}\Vert \partial_{i}(f-g)\Vert_{L^{2}(\mathbb{R}^{3})}\leq\varepsilon.$$

Now we have (using the notation $\Vert\cdot\Vert_{L^{2}(\mathbb{R}^{3})}:=\Vert\cdot\Vert_{2}$ from now on):

$$\bigg\Vert\frac{f}{\vert\cdot\vert}\bigg\Vert_{2}=\bigg\Vert\frac{f-g+g}{\vert\cdot\vert}\bigg\Vert_{2}\leq \bigg\Vert\frac{f-g}{\vert\cdot\vert}\bigg\Vert_{2}+\underbrace{\bigg\Vert\frac{g}{\vert\cdot\vert}\bigg\Vert_{2}}_{\leq 2 \Vert \nabla g\Vert_{2}}\leq \bigg\Vert\frac{f-g}{\vert\cdot\vert}\bigg\Vert_{2}+2(\Vert\nabla (f-g)\Vert_{2} + \Vert\nabla f\Vert_{2})$$

and hence

$$\bigg\Vert\frac{f}{\vert\cdot\vert}\bigg\Vert_{2}\leq 2\Vert\nabla f\Vert_{2} + 2\varepsilon+\bigg\Vert\frac{f-g}{\vert\cdot\vert}\bigg\Vert_{2}$$

First of all, is this right so far? Secondly, how can I proceed? If I somehow manage to relate $\bigg\Vert\frac{f-g}{\vert\cdot\vert}\bigg\Vert_{2}$ to $\Vert f-g\Vert_{2}\leq \varepsilon$ then I would recover the claimed inequality plus some $\varepsilon$, which would than prove the claim since this holds for all $\varepsilon>0$.

Thanks a lot!

$\endgroup$

1 Answer 1

1
$\begingroup$

The problem with your argument is that you need the claim to work for $f-g$ in order to proof that it works for $f$.

Instead, approximate $f$ by a sequence of smooth functions $(g_n)$. Then by the inequality for these smooth functions, one finds that $ \left( \frac{g_n}{|\cdot|} \right) $ is a Cauchy sequence in $L^2$. Then it remains to prove that the limit of the Cauchy sequence ($L^2$ is complete) is equal to $\frac{f}{|\cdot|}$.

$\endgroup$
1
  • $\begingroup$ Thanks for the answer. I try currently your proof idea. I have already proven that this is a Cauchy sequence. But how to I show that the limit is given by $f/\vert\cdot\vert$.? $\endgroup$
    – B.Hueber
    Nov 24, 2020 at 16:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .