product of $L^p$ functions is in $L^{p/2}$

Question

Let $p\ge 2$ and $\Omega\subseteq \mathbb{R}^n$ bounded, open. Let $f,g\in L^p(\Omega)$. Why is $fg\in L^{p/2}(\Omega)$?

I want to prove it using Hölder's inequlaity:

Set $\alpha=2=\frac{2p}{p}$ and $\alpha'=\alpha$. Then $\frac{1}{\alpha}+\frac{1}{\alpha'}=1 $ and $$\int|fg|^{p/2}dx\le\big(\int |f|^{ \frac{\alpha p}{2} } dx\big)^{1/\alpha}\big(\int |g|^{ \frac{\alpha' p}{2} } dx\big)^{1/\alpha'}=\big(\int |f|^{p} dx\big)^{1/2}\big(\int |g|^{ p } dx\big)^{1/2}$$ is not exactly that what I want to have, since it should be $$\int|fg|^{p/2}dx\le \big(\int |f|^{p} dx\big)^{1/p}\big(\int |g|^{ p } dx\big)^{1/p}.$$ How can I repair this, where is the mistake?

Answer 1

Your result is correct and enough. If $‖f‖_p =5$ is finite, so is $5^{p/2}$.

Answer 2

Some remarks.

• There is no mistake in your reasoning: you applied Cauchy-Schwarz inequality correctly. An alternative could be to use $2ab\leqslant a^2+b^2$ with $a=\left|f\right|^{p/2}$ and $b=\left|g\right|^{p/2}$.
• Usually, a way to check whether an inequality is correct is to replace a function by $\lambda$ times the function.If you do this with $f$ in the first inequality, you get $\lambda^{p/2}$ on both sides. If you do this in the last inequality you propose, you get $\lambda^{p/2}$ in the left hand side and $\lambda$ on the right hand side.