Assume that $f,g \in L_{10}(\mu)$. Prove that their product $fg$ belongs to $L_5(\mu)$.

Question

My question is this :

Assume that $f,g \in L_{10}(\mu)$. Prove that their product $fg$ belongs to $L_5(\mu)$. State any theorems, lemmas or propositions from the course that you use.

So my inital though was to use Holder's inequality, and we get

$\int |fg|d\mu = (\int|f|^{10}d\mu)^{\frac{1}{10}}(\int|g|^{10}d\mu)^{\frac{1}{10}}$ and then to take this to this to the power of five in which case we get

$(\int |fg|d\mu)^{5} = (\int|f|^{10}d\mu\int|g|^{10}d\mu)^{\frac{1}{2}}$, but I am struggling about where to go from here.

Help!

Answer 1

Let $f,g\in L_{10}(\mu)$ then by definition we have: $$\Big(\int|f|^{10}\,d\mu\Big)^{1/10}<\infty\Leftrightarrow\int|f|^{10}\,d\mu<\infty$$ Similarly $$\Big(\int|g|^{10}\,d\mu\Big)^{1/10}<\infty\Leftrightarrow\int|g|^{10}\,d\mu<\infty$$ Define $\widetilde{f}:=f^5$ and $\widetilde{g}:=g^5$. Now to show that $fg \in L_{5}(\mu)$ it is equivalent to show that $\widetilde{f}\widetilde{g}\in L_1(\mu)$ since $$\int|fg|^5\,d\mu<\infty\Leftrightarrow\int |\widetilde{f}\widetilde{g}|\,d\mu<\infty$$ On the other hand we have $f,g\in L_{10}(\mu)$ which is equivalent to $\widetilde{f},\widetilde{g}\in L_2(\mu)$. But then by Cauchy-Schwartz inequality you have $$\Big(\int|fg|^5\,d\mu\Big)^2=\Big(\int |\widetilde{f}\widetilde{g}|\,d\mu\Big)^2\leq\int |\widetilde{f}|^2\,d\mu\int|\widetilde{g}|^2\,d\mu<\infty$$ The result the follows.

Answer 2

You have $$|fg|^5=|f|^5 |g|^5\leq \frac{|f|^{10}+|g|^{10}}{2},$$ and integrating both sides gives you $$\int |fg|^5d\mu\leq \frac12\left(\int|f|^{10}d\mu+\int|g|^{10}d\mu\right)<+\infty,$$ hence $fg\in L^5(\mu)$.

Alternatively, you can apply Cauchy's inequality to $f^5$ and $g^5\in L^2(\mu)$; this is more related to your original attempt.