# FA6

Use the inequality $\alpha \beta \leq {\frac {\alpha ^{p}}{p}}+{\frac {\beta ^{q}}{q}}\,$ for $\alpha ,\beta \geq 0,{\frac {1}{p}}+{\frac {1}{q}}=1\,$ to prove that

$\int _{a}^{b}\left|f(t)g(t)\right|dt\leq \left(\int _{a}^{b}\left|f(t)\right|^{p}\right)^{{1/p}}\left(\int _{a}^{b}\left|g(t)\right|^{q}\right)^{{1/q}}\,$

for $p>1\,$ and ${\frac {1}{p}}+{\frac {1}{q}}=1\,$, and then use this to prove the triangular inequality:

$\left(\int _{a}^{b}\left|f(t)+g(t)\right|^{p}dt\right)^{{1/p}}\leq \left(\int _{a}^{b}\left|f(t)\right|^{p}\right)^{{1/p}}+\left(\int _{a}^{b}\left|g(t)\right|^{p}\right)^{{1/p}}\,$

## First Part

Let ${\hat {f}}\,$ and ${\hat {g}}\,$ be such that $\int _{a}^{b}|{\hat {f}}|^{p}dt=\int _{a}^{b}|{\hat {g}}|dt=1\,$.

Let $\alpha =|{\hat {f}}|\,$ and $\beta =|{\hat {g}}|\,$.

Using the inequality, $|{\hat {f}}{\hat {g}}|\leq |{\hat {f}}||{\hat {g}}|\leq {\frac {|{\hat {f}}|^{p}}{p}}+{\frac {|{\hat {g}}|^{q}}{q}}\,$.

Now integrate from $a\,$ to $b\,$.

$\int _{a}^{b}|{\hat {f}}{\hat {g}}|dt\leq \int _{a}^{b}{\frac {|{\hat {f}}|^{p}}{p}}dt+\int _{a}^{b}{\frac {|{\hat {g}}|^{q}}{q}}dt={\frac {1}{p}}+{\frac {1}{q}}=1\,$

Set ${\hat {f}}={\frac {f}{\left(\int _{a}^{b}|f|^{p}dt\right)^{{1/p}}}}\,$ and ${\hat {g}}={\frac {g}{\left(\int _{a}^{b}|g|^{q}dt\right)^{{1/q}}}}\,$.

Plug these into the previous equation and multiply by the denominator:

$\int _{a}^{b}|fg|dt\leq \left(\int _{a}^{b}|f|^{p}dt\right)^{{1/p}}\left(\int _{a}^{b}|g|^{q}dt\right)^{{1/q}}\,$

## Second Part

$|f+g|^{p}=|f+g||f+g|^{{p-1}}\,$

$\leq (|f|+|g|)|f+g|^{{p-1}}\,$

$|f||f+g|^{{p-1}}+|g||f+g|^{{p-1}}\,$

Therefore

$\int _{a}^{b}|f+g|^{p}dt\leq \int _{a}^{b}|f||f+g|^{{p+1}}dt+\int _{a}^{b}|g||f+g|^{{(p-1)q}}dt\,$

From the first part,

$\int _{a}^{b}|f||f+g|^{{p-1}}dt\leq \left(\int _{a}^{b}|f|^{p}dt\right)^{{1/p}}\left(\int _{a}^{b}|f+g|^{{(p-1)q}}\right)^{{1/q}}\,$

$\int _{a}^{b}|g||f+g|^{{p-1}}dt\leq \left(\int _{a}^{b}|g|^{p}dt\right)^{{1/p}}\left(\int _{a}^{b}|f+g|^{{(p-1)q}}\right)^{{1/q}}\,$

${\frac {1}{p}}+{\frac {1}{q}}=1\implies q+p=pq\implies (p-1)q=p\,$

$\int _{a}^{b}|f+g|^{p}dt\leq \left[\left(\int _{a}^{b}|f|^{p}dt\right)^{{1/p}}+\left(\int _{a}^{b}|g|^{p}dt\right)^{{1/p}}\right]\left(\int _{a}^{b}|f+g|^{p}dt\right)^{{1/q}}\,$
Multiply both sides by the inverse of the last term and use $1-{\frac {1}{q}}={\frac {1}{p}}\,$
$\left(\int _{a}^{b}|f+g|^{p}dt\right)^{{1/p}}\leq \left(\int _{a}^{b}|f|^{p}dt\right)^{{1/p}}+\left(\int _{a}^{b}|g|^{p}dt\right)^{{1/p}}\,$