Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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:… – 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
up vote 1 down vote accepted

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).

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.