Proof of $\big|\int_E (f\cdot g)(x) \text{d}x\big| \leq [\int_E |f(x)|^p\text{d}x]^{(1/p)} \cdot [\int_E |g(x)| ^q\text{d}x]^{(1/q)}$

Question

The answer to this question is given in here, but I cannot fill the gaps between the accepted answer, so here is what I have done with the guide of @Raito.

My work:

By Young's inequality, we do know that

$$|f\cdot g| \leq \frac{|f|^p}{p} + \frac{ |g|^q}{ q} = \frac{\lambda^p |f|^p}{p} + \frac{ |g|^q}{\lambda^q q}.$$ Now choose $\lambda = \left(\dfrac{|g|^q }{|f|^p }\right)^{\frac{1 }{p+q } },$ then we have $$= |f| \cdot |g|.$$

Now, we also do know that

$$\bigg|\int_E f\cdot g \ \text{d}x\bigg| \leq \int_E |f\cdot g| \text{d}x \leq \int_E \frac{|f|^p}{p} \text{d}x + \int_E \frac{ |g|^q}{ q} \text{d}x \leq \int_E |f| \text{d}x \cdot \int_E |g| \text{d}x $$

But after that I couldn't figure out how to get the $1/p$ and $1/q$ powers of those integrals.

Answer 1

Let $\widetilde{f}=\dfrac{|f|}{\|f\|_{p}}$, $\widetilde{g}=\dfrac{|g|}{\|g\|_{q}}$, then $\|\widetilde{f}\|_{p}=\|\widetilde{g}\|_{q}=1$. And \begin{align*} \widetilde{f}\widetilde{g}&\leq\dfrac{\widetilde{f}^{p}}{p}+\dfrac{\widetilde{g}^{q}}{q},\\ \int\widetilde{f}\widetilde{g}dx&\leq\dfrac{1}{p}\int\widetilde{f}^{p}dx+\dfrac{1}{q}\int\widetilde{g}^{q}dx\\ \int\widetilde{f}\widetilde{g}dx&\leq\dfrac{1}{p}\|\widetilde{f}\|_{p}+\dfrac{1}{q}\|\widetilde{g}\|_{q}\\ \int\widetilde{f}\widetilde{g}dx&\leq\dfrac{1}{p}+\dfrac{1}{q}, \end{align*} so \begin{align*} \int\widetilde{f}\widetilde{g}dx&\leq 1,\\ \int|fg|dx&\leq\|f\|_{p}\|g\|_{q}. \end{align*}