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In Royden's real analysis, the proof for the Hölder inequality (on pg. 121) is stated as follows:

If $p$ and $q$ are nonnegative extended real numbers such that

$\frac{1}{p} + \frac{1}{q} = 1,$

and if $f \in L^p$ and $g \in L^q$, then $f \cdot g \in L^1$ and

$\int |fg| \leq ||f||_p \cdot ||g_q||.$

The proof is trivial for $p=\infty$ or $q = \infty$ so assume $1 < p < \infty$ and $1 < q < \infty$.

In the proof of this, the function $h(x) = g(x)^{q-1} = g(x)^{\frac{q}{p}}$ is defined. This yields $g(x) = h(x)^{\frac{p}{q}}$.

After defining $h$, the book says, without explanation,

$ptf(x)g(x) = ptf(x)h(x)^{p-1} \leq (h(x)+tf(x))^p - h(x)^p.$

Where does this inequality come from? I want to say that somehow it involves convexity, but I am not sure.

  • 0
    The inequality is an application of Lemma 3 on the same page. In the Lemma, $t$ is non-negative.2011-11-17

2 Answers 2

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You can see this by the mean value theorem, applied to $\phi(s)=s^p$: \phi(h+tf)-\phi(h) = \phi'(h+\theta )tf, where $\theta$ is between $0$ and $tf$. Just notice that \phi'(h+\theta )tf\geq\phi'(h)tf, which comes from the fact that the derivative of $\phi$ is increasing when $1 (which is equivalent to convexity).

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Since $p\geq 1$, you can apply Bernoulli's inequality to obtain (for y,z> 0) $(y+z)^p=(1+z/y)^p y^p\geq (1+p(z/y))\, y^p=y^p+pzy^{p-1}.$ The inequality is also true when $y,z\geq 0$.

Royden says that we only need to consider the case when $f\geq 0 $ and $g\geq 0$, so plug in $y=h(x)$ and $z=t f(x)$ and you're done.