$\|f\|_{L^{3}(\mathbb R)}^{3} \leq C \|f\|_{L^{2}(\mathbb R)} \|\nabla f\|_{L^{2}(\mathbb R)}^{3}$ ; for some constant $C$? 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,
 A: 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 $f\in H^1(\mathbb R)$.
