Let $f: [0,\infty) \rightarrow [0,\infty)$ be an onto, strictly increasing function. This problem looks like $\textbf{Young's Inequality}$ but it does not say that $f(0)=0$ or does the fact that I am told that the function is strictly increasing imply that $f(0)=0$.
That is the doubt I have with this problem. Thank you very much in advance to anyone that can help me clear that doubt!
$f(0) = 0$ is implied by the conditions: Assume that $f(0) = b >0$. Then $f([0, \infty)) \subset [b, \infty) $ because $f$ is increasing. On the other hand, $0 \in f([0, \infty))$ because $f$ maps onto $[0, \infty)$.
Or shorter: $$ 0 \le f(0) \le f(f^{-1}(0)) = 0 \implies f(0) = 0 \, . $$
So you can apply Young's inequality. For a proof see Proof of Young's inequality or Prove that for $a,b > 0$, we have $\int_{0}^a f(x)dx+\int_{0}^b f^{-1}(x)dx \geq ab$.
