Prove that $fg\in L^r(\Omega)$ if $f\in L^p(\Omega),g\in L^q(\Omega)$, and $\frac1 p+\frac1 q=\frac1 r$

Question

Can anyone give me a hint for proving the following:

Let $\Omega$ be a measure space. Assume $f \in L^p(\Omega)$ and $g \in L^q(\Omega)$ with $1 \leq p, q \leq \infty$ and $\frac1p + \frac1q \leq 1$. Prove that $fg \in L^r(\Omega)$ with $\frac1r = \frac1p + \frac1q$.

Note: One should be able to use (the standard) Hölder inequality. Notice that if you have $\frac1p + \frac1q = 1$ you recover the former result.

Answer 1

Using the standard Hölder inequality: $$ \|fg\|_r^r=\|f^rg^r\|_1\le\|f^r\|_{p/r}\|g^r\|_{q/r}=\|f\|_p^r\|g\|_q^r $$ Since $\frac{r}{p}+\frac{r}{q}=1$.

Answer 2

Hint: $$ 1 = \frac{r}{p}+\frac{r}{q} = \frac{1}{\frac{p}{r}}+\frac{1}{\frac{q}{r}} $$

Answer 3

we can remark that : $$\frac 1{p'} + \frac 1{q'} = 1$$ where : $$p'=\frac pr ; q'=\frac qr$$ and proof that $f^r \in L^{p'}$ and $g^r \in L^{q'}$