Hölder’s Inequality

If $a = 0$ or $b = 0$ both sides are trivial ($0 \le$ the right side), so take $a, b > 0$.

Then $ab = \exp(\log a + \log b) = \exp\left(\frac{1}{p}\log(a^p) + \frac{1}{q}\log(b^q)\right)$.

Since $\log$ is concave, the weighted Jensen inequality $\log\left(\frac{X}{p} + \frac{Y}{q}\right) \ge \frac{1}{p}\log X + \frac{1}{q}\log Y$ ($X, Y > 0$) holds.

With $X = a^p$, $Y = b^q$ this gives $\frac{1}{p}\log(a^p) + \frac{1}{q}\log(b^q) \le \log\left(\frac{a^p}{p} + \frac{b^q}{q}\right)$. Applying the monotone $\exp$ to both sides, $\exp\left(\frac{1}{p}\log(a^p) + \frac{1}{q}\log(b^q)\right) = ab \le \frac{a^p}{p} + \frac{b^q}{q}$ — Young's inequality proved.

Equality case: $\log$ is strictly concave, so for $a, b > 0$ Jensen gives equality only when $X = Y$, i.e. $a^p = b^q$.

Verify the zero cases directly as well: if $a = 0$ then the left side is $ab = 0$ and the inequality reads $0 \le \frac{b^q}{q}$, with equality only when $b^q = 0$, i.e. $b = 0$ — exactly what $a^p = b^q$ (namely $0 = b^q$) asserts; the case $b = 0$ is symmetric.

Hence in all cases equality holds if and only if $a^p = b^q$.

This equality condition becomes Hölder's equality case in problem 3 below.

If $\|a\|_p = 0$ then every $a_k = 0$ and both sides are 0; likewise for $\|b\|_q = 0$.

So assume $\|a\|_p, \|b\|_q > 0$.

Define the normalized sequences $A_k = |a_k|/\|a\|_p$, $B_k = |b_k|/\|b\|_q$; then $\sum A_k^p = 1$ and $\sum B_k^q = 1$.

Apply Young's inequality to each $k$: $A_k B_k \le \frac{A_k^p}{p} + \frac{B_k^q}{q}$. Summing over $k = 1, \dots, n$ gives $\sum A_k B_k \le \frac{1}{p}\sum A_k^p + \frac{1}{q}\sum B_k^q = \frac{1}{p} + \frac{1}{q} = 1$.

Multiplying both sides by $\|a\|_p\|b\|_q$ gives $\sum |a_k b_k| = \|a\|_p\|b\|_q \sum A_k B_k \le \|a\|_p\|b\|_q$ — Hölder's inequality proved.

The case $p = q = 2$ is Cauchy–Schwarz.

Sliding $p$ in the scene shows how this right side inflates and deflates.

Separate the degenerate case first.

If $\|a\|_p \|b\|_q = 0$ then one of $a, b$ is the zero sequence, so $\sum |a_k b_k| = 0$ and the right side is also $0$ — equality holds automatically.

Note: with $b = 0$ but $a \ne 0$, this equality holds, yet no constant — not even $C \ge 0$, let alone $C > 0$ — satisfies $|a_k|^p = C|b_k|^q$ (the right side is identically 0 while some $|a_k|^p > 0$).

That is why the first branch of the statement is indispensable.

In both directions below, the degenerate case is settled by this paragraph, and the proportional branch is treated under $\|a\|_p, \|b\|_q > 0$.

$(\Rightarrow)$ Assume $\|a\|_p, \|b\|_q > 0$ and that equality holds overall.

In problem 2's proof the inequality happened in exactly one place: Young's inequality $A_k B_k \le \frac{A_k^p}{p} + \frac{B_k^q}{q}$ applied to each term.

Equality gives $\sum_k A_k B_k = 1$, while $\sum_k \left(\frac{A_k^p}{p} + \frac{B_k^q}{q}\right) = \frac{1}{p}\sum A_k^p + \frac{1}{q}\sum B_k^q = 1$ is an identity, so if any single term were strict, the summed inequality would be strict — hence equality holds for every $k$.

By problem 1's equality condition, $A_k^p = B_k^q$ for every $k$; unwinding the normalization gives $\frac{|a_k|^p}{\|a\|_p^p} = \frac{|b_k|^q}{\|b\|_q^q}$ — i.e. $|a_k|^p = C|b_k|^q$ with $C = \|a\|_p^p/\|b\|_q^q > 0$ (both norms are positive) independent of $k$.

$(\Leftarrow)$ Conversely, assume one of the two branches.

If the degenerate branch $\|a\|_p \|b\|_q = 0$ holds, the first paragraph already verified equality with both sides $0$ — this also covers the $C = 0$ case: if $|a_k|^p = 0 \cdot |b_k|^q$ for all $k$, then $a = 0$ and the first branch applies with no division by any norm.

For the proportional branch, assume $|a_k|^p = C|b_k|^q$ for all $k$ and some $C > 0$.

If $\|b\|_q = 0$ then $b = 0$ and proportionality forces $a = 0$ as well, returning to the degenerate branch; if $\|a\|_p = 0$ then $a = 0$, again the degenerate branch.

In the remaining case $\|a\|_p, \|b\|_q > 0$, division by the norms is legitimate: summing the proportionality over $k$ forces $C = \|a\|_p^p/\|b\|_q^q$, and after normalization $A_k^p = \frac{|a_k|^p}{\|a\|_p^p} = \frac{|b_k|^q}{\|b\|_q^q} = B_k^q$, so by problem 1's equality condition every Young's inequality is an equality: $A_k B_k = \frac{A_k^p}{p} + \frac{B_k^q}{q}$.

Summing over $k$ gives $\sum_k A_k B_k = \frac{1}{p}\sum_k A_k^p + \frac{1}{q}\sum_k B_k^q = \frac{1}{p} + \frac{1}{q} = 1$, i.e. $\sum_k |a_k b_k| = \|a\|_p \|b\|_q$ — equality in Hölder's inequality.

In the scene, $a_k = 1/k$ and $b_k = 1$ have both norms positive but $|a_k|^p$ is not constant, so the gap can never be 0 — the gap closing as $p \to 1^+$ is a collapse of the right side itself, not proportionality.