Deducing properties of the $\ell_3$ norm from the $\ell_1$ and $\ell_2$ norms

Question

Suppose we have a function $f: [0,1] \rightarrow \mathbb{R}$, $f(x) \geq 0$ normalised so that $\|f\|_1 = 1$, where $$ \| f\|_p = \left( \int_0^1 f(x)^p d x \right)^{1/p}. $$ Moreover, we know that $\| f \|_2 = A$. Is it possible to bound $\| f \|_3$ in terms of $A$? What about if we know, in addition, $f$ is bounded as $f \leq B$?

Answer 1

Let $f(x)=\frac{1}{\sqrt[3]{x}}$ for $x>0$ and $f(0)=0$. Now, $||f||_3=\infty$, but $$||f||_1=\int_0^1\frac{1}{\sqrt[3]{x}}\ dx=\frac{3}{2}, $$$$||f||_2^2=\int_0^1\frac{1}{x^{2/3}}\ dx=3.$$ That is above integrals are finite, normalise $f$ by multiplying $2/3$. So, first problem has negative answer.

Now, for the second problem has affirmative answer. Note that $$\int_0^1|f|^3=\int_0^1|f|^2|f|\leq \min\bigg\{B^2\int_0^1|f|,\ B\int_0^1|f|^2\bigg\}<\infty$$