$\|f\|_{L^{3}(\mathbb R)}^{4} \leq C \|f\|_{L^{2}(\mathbb R)}^{3} \|\nabla f\|_{L^{2}(\mathbb R)}$ ; for some constant $C$? Let $f\in H^{1}(\mathbb R).$ (Sobolev space)

My Question: Is it true that:
  $\|f\|_{L^{3}(\mathbb R)}^{4} \leq C \|f\|_{L^{2}(\mathbb R)}^{3} \|\nabla f\|_{L^{2}(\mathbb R)}$ ; for some constant $C$? If yes, How to prove it?

 A: The inequality is wrong. Take the function 
$$ \left\lbrace \begin{array} .\phi(x) = x+2 & if & x \in [-2,-1] \\
\phi(x) = 1 & if & x \in [-1,1] \\
 \phi(x) = 2-x & if & x \in [1,2] \\
 \phi(x) = 0 & if & x \not\in [-2,2] 
\end{array} \right. $$
$\phi$ is in $H^1(\mathbb{R})$. Now consider the sequence of functions $\phi_n(x) = \phi(\frac{x}{n})$
You have :
$$\| \phi_n \|_3^3 = 2n + 2 \int_n^{2n} |2-\frac{x}{n} |^3 dx = \frac{5}{2}n$$
$$\| \phi_n \|_2^2 = 2n + 2 \int_n^{2n} |2-\frac{x}{n} |^2 dx = \frac{8}{3}n$$
$$\| \phi_n' \|_2^2 = 2 \int_n^{2n} |\frac{1}{n} |^2 dx = \frac{2}{n}$$
If the inequality was true, you would have for all $n$
$$(\frac{5}{2}n)^{\frac{4}{3}} \leq C (\frac{8}{3}n)^{\frac{3}{2}}(\frac{2}{n})^{\frac{1}{2}}$$
So you would have for all $n$
$$n^{\frac{4}{3}} \leq C n$$
$$n^{\frac{1}{3}} \leq C$$
A: Interpolation inequalities are becoming a recurrent topic nowadays! Cauchy-Schwarz and integration by parts are the key tools, as always. For any $x>0$ we have:
$$ f(x)^2 = f(0)^2+2\int_{0}^{x}f'(y)\,f(y)\,dy \tag{1} $$
where $f(0)^2$ has to be intended as the Lebesgue-principal value in zero, i.e.:
$$ f(0)^2 = \lim_{h\to 0^+}\frac{1}{2h}\int_{-h}^{h}f(x)^2\,dx$$
since we are not dealing with pointwise-defined functions, but that does not matter so much. 
By $(1)$, it follows that $f$ belongs to $L^\infty(\mathbb{R})$, since by the Cauchy-Schwarz inequality we have:
$$ f(x)^2 \leq f(0)^2 + 2\,\|f\|_2\,\|f'\|_2.\tag{2}$$
Interpolating between $L^2$ and $L^\infty$ through $(2)$, we have:
$$ \| f\|_3^3 \leq \|f\|_2^2 \sqrt{f(0)^2+2\,\|f\|_2\,\|f'\|_2}. \tag{3}$$
Assuming $f(0)^2=0$ - or just rewriting $(1)$ as $f(x)^2=2\int_{-\infty}^{x}f'(y)\,f(y)\,dy$ - it follows that:
$$ \|f\|_3^4 \leq 2^{2/3}\,\|f\|_2^{10/3}\,\|f'\|_2^{2/3}\tag{4} $$
Assuming that $f$ is compact supported, now it is possible to weaken the previous inequality by bounding $\|f\|_2^{1/3}$ with $\left(C\,\|f'\|_2\right)^{1/3}$ through Wirtinger's inequality, proving the claim. 
Otherwise, the claim may not hold, as shown by Tryss.
