Valid proof of Young's inequality?

Solution 1:

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.