# Hölder Inequality proof in Royden

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.

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what is $t$ in the formula? –  Olivier Bégassat Sep 20 '11 at 20:54
The book is not explicit about what $t$ is, but later on in the proof they differentiate with respect to $t$ and then set $t=0$. So I interpret $t$ as just some independent variable, which can vary seemingly anywhere. –  tomcuchta Sep 20 '11 at 20:55
This should follow from the binomial theorem if $h$ and $f$ are nonnegative. Does Royden assume, $f, h \geq 0$ at this point in the proof? For more, see: en.wikipedia.org/wiki/… –  JavaMan Sep 20 '11 at 21:52
The inequality is an application of Lemma 3 on the same page. In the Lemma, $t$ is non-negative. –  Nana Nov 17 '11 at 4:46

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 $1<p$ (which is equivalent to convexity).
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.