Functional inequalities involving the condition $\left(\int_0^t f(x)dx\right)^2 \ge \int_0^t f(x)^3dx$

Question

I was reading the solution to a functional inequality in an article when the author made the following remark without giving any proof: let $f(x): [0, \infty]\to[0, \infty]$ be locally integrable and such that $$\left(\int_0^t f(x)dx\right)^2 \ge \int_0^t f(x)^3dx$$ for all $t>0$. Then, the following statement is true:

$\int_0^t f(x)^\gamma dx \le \frac{1}{\gamma +1}\left(2\int_0^t f(x)dx\right)^{(\gamma + 1)/2}$ for all positive $t$ and $\gamma \in [1,3]$.

Again, there is no proof in the article, so I don't know if this is fairly easy to prove or very involved. One thing that might be worth mentioning is that the inequalities above become exact when $f(x)=x$. I am wondering if anyone has an idea or have seen these before.

Answer 1

I don't know about the precise multiplicative prefactor $\frac{2^{\frac{\gamma+1}2}}{\gamma+1}$, but I can tell you how to get $$ \int _0^t f^\gamma\leq \left(\int_0^t f\right)^{\frac{\gamma+1}{2}}. $$ This is a simple interpolation inequality: for any fixed $\gamma\in (1,3)$ (the boudnary cases $\gamma=1,3$ are immediate) there is a unique $\theta=\theta(\gamma)\in(0,1)$ such that $$ \frac 1\gamma=\theta\frac{1}{1}+(1-\theta)\frac{1}{3}. $$ By standard interpolation you get $$ \|f\|_{\gamma}\leq \|f\|_{1}^{\theta} \|f\|_{3}^{1-\theta}. $$ Your assumption is equivalent to the "$L^3$ by $L^1$ control" $\|f\|_3\leq \|f\|_1^{\frac{2}{3}}$, which then immediately implies $$ \|f\|_{\gamma}\leq \|f\|_{1}^{\theta+\frac 23(1-\theta)}. $$ Leveraging the explicit value of $\theta=\theta(\gamma)$, some straightforward algebra then finally gives the exact exponent $\theta+\frac 23(1-\theta)=\frac{\gamma+1}{2}$.