FA6

From Exampleproblems

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

$\int_a^b \left| f(t)g(t) \right| dt \le \left( \int_a^b \left| f(t) \right|^p \right)^{1/p} \left( \int_a^b \left| g(t) \right|^q \right)^{1/q}\,$

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

$\left( \int_a^b \left| f(t)+g(t) \right|^p dt \right)^{1/p} \le \left( \int_a^b \left| f(t) \right|^p \right)^{1/p} + \left( \int_a^b \left| g(t) \right|^p \right)^{1/p}\,$

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, $|\hat{f}\hat{g}| \le |\hat{f}||\hat{g}| \le \frac{|\hat{f}|^p}{p}+\frac{|\hat{g}|^q}{q}\,$ .

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

$\int_a^b |\hat{f}\hat{g}| dt \le \int_a^b \frac{|\hat{f}|^p}{p} dt + \int_a^b \frac{|\hat{g}|^q}{q} dt = \frac{1}{p} + \frac{1}{q} = 1\,$

Set $\hat{f}=\frac{f}{\left(\int_a^b |f|^pdt\right)^{1/p}}\,$ and $\hat{g}=\frac{g}{\left(\int_a^b |g|^qdt\right)^{1/q}}\,$ .

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

$\int_a^b |fg| dt \le \left( \int_a^b |f|^p dt \right)^{1/p} \left(\int_a^b |g|^q dt\right)^{1/q}\,$

Second Part

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

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

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

Therefore

$\int_a^b |f+g|^p dt \le \int_a^b |f||f+g|^{p+1} dt + \int_a^b |g||f+g|^{(p-1)q} dt\,$

From the first part,

$\int_a^b |f||f+g|^{p-1} dt \le \left(\int_a^b |f|^p dt\right)^{1/p} \left(\int_a^b|f+g|^{(p-1)q} \right)^{1/q}\,$

$\int_a^b |g||f+g|^{p-1} dt \le \left(\int_a^b |g|^pdt\right)^{1/p} \left(\int_a^b|f+g|^{(p-1)q} \right)^{1/q}\,$

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

Adding these two integrals,

$\int_a^b |f+g|^p dt \le \left[ \left( \int_a^b|f|^p dt\right)^{1/p} + \left( \int_a^b|g|^p dt\right)^{1/p}\right] \left( \int_a^b |f+g|^p dt\right)^{1/q} \,$

Multiply both sides by the inverse of the last term and use $1-\frac{1}{q}=\frac{1}{p}\,$

$\left(\int_a^b |f+g|^p dt\right)^{1/p} \le \left( \int_a^b |f|^p dt \right)^{1/p} + \left( \int_a^b |g|^pdt\right)^{1/p}\,$

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FA6

From Exampleproblems

First Part

Second Part

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