Hölder's inequality

Question

Let p and q be positive real numbers such that $\frac{1}{p}$ + $\frac{1}{q}$ = 1

Here is the Hölder's inequality:

|$\int_a^bfg$ $d\alpha$| $\leq$ {$\int_a^b$ $|f|^p$ $d\alpha$}$^{1/p}$ {$\int_a^b$ $|g|^q$ $d\alpha$}$^{1/q}$

I've a trouble to proof it.

I've tried to proof from this inequality:

f(x)g(x) $\leq$ $\frac {f(x)^p}{p}$ + $\frac {g(x)^q}{q}$

Any help would be appreciated.

Answer 1

Hint

I recall that $\|h\|_{L^r}:=\left(\int_a^b|h|^r\right)^{1/r}$. Taking $$u(x)=\frac{|f(x)|}{\|f\|_{L^p}}\quad \text{and}\quad v(x)=\frac{|g(x)|}{\|g\|_{L^q}},$$ in Young's inequality gives $$\frac{|f(x)|}{\|f\|_{L^p}}\cdot \frac{|g(x)|}{\|g\|_{L^q}}\leq \frac{1}{p}\cdot \frac{|f(x)|^p}{\|f\|_{L^p}^p}+\frac{1}{q}\cdot \frac{|g(x)|^q}{\|g\|_{L^q}^q}.$$

Integrating both side gives the wished result.