If $f\in L^4[0,1]$ and $\|f\|_4\leq C\|f\|_2$, then $\|f\|_2\leq C_1\|f\|_1$

I'm trying to solve the following problem. Let $\|f\|_p=\left(\int_0^1|f|^p\right)^{1/p}$ be the usual $L^p$ norm.

Let $f\in L^4[0,1]$ and suppose there is a constant $C$ such that $\|f\|_4\leq C\|f\|_2$. Prove that there exists a constant $C_1$ depending only on $C$ such that $\|f\|_2\leq C_1\|f\|_1$.

So far I can prove that $\|f\|_1\leq\|f\|_2\leq \|f\|_4$ using Holder's inequality. How do I proceed? Any hints regarding this are greatly appreciated. Thanks.

Solution 1:

Write $$ \int|f|^2=\int|f|^{4/3}|f|^{2/3}, $$ and apply the Hölder inequality with the exponents $p=3$ and $q=\frac32$.

Solution 2:

First note that the inequality will hold for any $C_1 \geq 0$ when $\|f\|_2 = 0$, so we may assume $\|f\|_2 \neq 0$.

By Holder's inequality, we have $$\|f\|_2^2 = \||f|^2\|_1 = \||f|^{2/3}|f|^{4/3}\|_1 \leq \||f|^{2/3}\|_{3/2}\||f|^{4/3}\|_3 = \|f\|_1^{2/3}\|f\|_4^{4/3}$$

Since $\|f\|_4 \leq C \|f\|_2$, this gives $$\|f\|_2^2 \leq C^{4/3}\|f\|_1^{2/3}\|f\|_2^{4/3}$$

As you had noted: for measurable functions $g : [0,1] \to \mathbb{R}$, $\|g\|_2 \leq \|g\|_4$, so in particular, $\|f\|_2 < \infty$. Therefore, by dividing by $\|f\|_2^{4/3} > 0$ we have $$\|f\|_2^{2/3} \leq C^{4/3}\|f\|_1^{2/3}$$

Finally, since $x \mapsto x^{3/2}$ is monotone increasing on $x \geq 0$, $$\|f\|_2 \leq C^2\|f\|_1$$