13
$\begingroup$

Part of an exercise to prove Holder's inequality in Rudin involves proving Young's Inequality... That is, given $\frac{1}{p}+\frac{1}{q} = 1$, prove $$ab \leqslant \frac{a^p}{p} + \frac{b^q}{q}.$$

Here's my attempt at a proof:

Let $$f(x) = \frac{x^p}{p} + \frac{b^q}{q} -bx$$
then, $$f'(x) = x^{p-1} -b$$ so that $f$ attains a minimum at $$x=b^{\frac{1}{p-1}}.$$ Since $\frac{p}{p-1} = q$, this is equivalent to saying that $f$ attains its minimum when $x^p = b^q$. Now, we need show that $$f(b^{\frac{1}{p-1}}) = 0.$$ We compute
$$ \begin{align} f(b^{\frac{1}{p-1}}) &= \frac{(b^{\frac{1}{p-1}})^p}{p} + \frac{b^q}{q} - bb^{\frac{1}{p-1}} \\ &= \frac{b^q}{p} + \frac{b^q}{q} - b^{\frac{1}{p-1} +1} \\ &= \frac{b^q(p+q)}{pq} - b^{\frac{1}{p-1} +1} \\ &= b^q - b^{\frac{1}{p-1} +1} \\ &= b^q - b^q \\ & = 0 \end{align}$$ where $b^{\frac{1}{p-1} +1} = b^q$ since $$\frac{1}{p-1} +1 = \frac{1}{p-1} +\frac{p-1}{p-1} = \frac{p}{p-1} = q.$$ Thus, $f(x) = 0$ only when $x^p = b^q$. This is the global minimum of $f$ since $f^{''} \geq 0$ and analysis of concavity. Therefore, if $x > b^{\frac{1}{p-1}}$, $f(x) > 0$. That is, if $x^p > b^q$, the inequality holds. A similar analysis for $g(y) = \frac{a^p}{p} + \frac{y^q}{q} -ay$ shows that $g(y) > 0$ if $y^q > a^p$. Combining these two statements yields that, if $a^p \neq b^q$, the inequality holds, so we're done.

Is this a valid proof?

If so, if anyone could provide any alternative proofs, I'd be more than interested to see them.

$\endgroup$
3
  • 2
    $\begingroup$ Excellent! The standard proof is to use the concavity of the log function and show the inequality almost immediately. But this is creative! $\endgroup$
    – Mark Viola
    Mar 31, 2015 at 4:28
  • 1
    $\begingroup$ I think that (inequality) would be suitable for this question. (Probably the most suitable.) However, you have already used maximum of 5 tags. $\endgroup$ Mar 31, 2015 at 7:16
  • $\begingroup$ @Martin Sleziak changed accordingly. $\endgroup$ Mar 31, 2015 at 7:17

1 Answer 1

4
$\begingroup$

As mentioned above, the standard proof uses concavity. You said you were interested in other proofs so here is how I learned it:

We will first assume $a,b \ne 0$ as this then becomes trivial. An equivalent statement to the one we want to prove is the following:

$$\frac{a}{b^{q-1}} \le \frac{1}{p} \cdot \frac{a^p}{b^q} + \frac{1}{q}$$

(divide Young's Inequality by $b^q$).

Let $t = \frac{a^p}{b^q}$. Then $t^{\frac{1}{p}} = \frac{a}{b^{q-1}}$, since $\frac{q}{p} = q - 1$ , we want to show that

$$t^{\frac{1}{p}} - \frac{1}{p} t \le \frac{1}{q}$$

Let $f(t) = t^{\frac{1}{p}} - \frac{1}{p} t$. Then $f(1) = 1 - \frac{1}{p} = \frac{1}{q}$. If we can show that this is the maximum of $f(t)$ when $t > 0$, we are done.

$$f'(t) = \frac{1}{p} t^{\frac{1}{p} - 1} - \frac{1}{p} = \frac{1}{p} \left(t^{-\frac{1}{q}} - 1\right)$$

When $0 < t < 1$, we see that $f'(t) > 0$ and similarly when $t > 1$, we see that $f'(t) < 0$. Thus, $\frac{1}{q}$ is a local max of $f(t)$ and so we have shown that

$$t^{\frac{1}{p}} - \frac{1}{p} t \le \frac{1}{q}$$

Now substituting back in $t = \frac{a^p}{b^q}$ we get Young's Inequality to come out with some algebraic manipulations.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .