# A problem on the bounds of Lp-norms

Let $L>0$ and $\Omega$ be the set of all integrable functions from $[0,L]$ to $[0,+\infty]$. Also, Let $f\in \Omega$ such that $\left \| f \right \|_{1}=1$. Find the tightest possible bounds for:
\begin{align}\left \| f \right \|_{2}\times \left \| \frac{1}{f} \right \|_{2}\end{align}

• What have you tried? The tightest lower bound is immediate from a very fundamental inequality on $L^p$ spaces... – Anthony Carapetis Aug 23 '13 at 6:21
• @AnthonyCarapetis I have tried Holder inequality. But I am not very familiar with the reverse Holder inequality, which might be needed for this kind of problems. After all, I preferred to ask those who are more skillful in these areas. Can you solve it? – Amir Kazemi Aug 23 '13 at 6:26
• Given your hypotheses as written, I'm unsure you're forbidden from taking $f=(2/L)\chi_{[0,L/2]}$. Then $f$ is integrable on $[0,L]$, and has unit $L^1$-norm -- satisfying your hypotheses. However, $f$ has positive $L^2$-norm and $1/f$ has infinite $L^2$ norm. – Adam Azzam Aug 23 '13 at 7:07
• @A.AdamAzzam Thanks for your reply. Actually my original problem is: math.stackexchange.com/q/457500/88652. In this link, I would appreciate it if you could help me to find a lower- and upper-bound, if it is not possible to find the extrema. – Amir Kazemi Aug 23 '13 at 9:36

Hint: The information about $\|f\|_1$ is a red herring.
• There is no finite upper bound. The lower bound does not depend on $\|f\|_1$ at all, because $f$ and $c f$ give the same result for any positive constant $c$. – Robert Israel Aug 23 '13 at 14:19
For the upper bound, you can try setting it up with Hölder's inequality; but you can't get the required indices to stay $\ge 1$. You should try looking at some families of simple examples to see if such a bound is possible - in particular note that we can make $\Vert f \Vert$ very large while keeping $\Vert f^{-1} \Vert$ fairly steady if the function is highly peaked in a small region.