$\|f\|_{L^{3}(\mathbb R)}^{3} \leq C \|f\|_{L^{2}(\mathbb R)} \|\nabla f\|_{L^{2}(\mathbb R)}^{3}$ ; for some constant $C$?

Question

Let $f\in H^{1}(\mathbb R)$ (Sobolev space).

My Question: Is it true that: $\|f\|_{L^{3}(\mathbb R)}^{3} \leq C \|f\|_{L^{2}(\mathbb R)} \|\nabla f\|_{L^{2}(\mathbb R)}^{3}$ ; for some constant $C$? [If it is true, then I guess some where I have to use Sobolev inequality but I don't know all kinds of Sobolev inequality]

Thanks,

Answer 1

There is no such $C$. Replace $f$ with $af$, $a>0$, and then you would have $$ a^3\|f\|_{L^{3}(\mathbb R)}^{3} \leq a^4C \|f\|_{L^{2}(\mathbb R)} \|\nabla f\|_{L^{2}(\mathbb R)}^{3} $$ or $$ \|f\|_{L^{3}(\mathbb R)}^{3} \leq aC \|f\|_{L^{2}(\mathbb R)} \|\nabla f\|_{L^{2}(\mathbb R)}^{3}. $$ Then take $a\to 0$ and obtain a contradiction.

However, $$ \|\,f\|_{L^{3}(\mathbb R)}^{6} \leq 2\, \|\,f\|_{L^{2}(\mathbb R)}^5 \|\nabla f\|_{L^{2}(\mathbb R)}, \tag{1} $$ is indeed correct!

Explanation. Let $f\in C_0^\infty(\mathbb R)$, then $$ f(x)=\int_{-\infty}^xf'(s)\,ds,\quad\text{hence}\,\,\, \lvert\,f(x)\rvert\le\int_{-\infty}^x\lvert\, f'(s)\rvert\,ds\le\int_{-\infty}^\infty\lvert\, f'(s)\rvert\,ds=\|\,f'\|_{L^1} $$ and thus $\,\|\,f\|_{L^\infty}\le \|\,f'\|_{L^1}$. In particular, $$ \|\,f\|_{L^\infty}^2=\|\,f^2\|_{L^\infty}\le\|\,(f^2)'\|_{L^1}=2\|\,ff'\|_{L^1}\le2\|\,f\|_{L^2}\|\,f'\|_{L^2}, $$ and hence $$ \|\,f\|_{L^\infty}\le 2^{1/2}\|\,f\|_{L^2}^{1/2}\|\,f'\|_{L^2}^{1/2}. $$ Next observe that $$ \|\,f\|_{L^3}^3=\int_{-\infty}^\infty \lvert\,f(x)\rvert^3\,dx\le \|\,f\|_{L^\infty}\int_{-\infty}^\infty \lvert\,f(x)\rvert^2\,dx\le 2^{1/2}\|\,f\|_{L^2}^{5/2}\|\,f'\|_{L^2}^{1/2}, $$ which implies $(1)$, as $C_0^\infty(\mathbb R)$ is dense in $H^1(\mathbb R)$.

In general, if $2<p\le \infty$, then there exists a $c_p>0$, such that $$ \|\,f\|_{L^p}^{2p}\le c_p\|\,f\|_{L^2}^{p+2}\|\,f'\|_{L^2}^{p-2}, $$ for every $f\in H^1(\mathbb R)$.