Hölder’s Inequality

For conjugate exponents $1/p + 1/q = 1$ ( $1 < p, q < \infty$ ), prove Young's inequality $ab \le \frac{a^p}{p} + \frac{b^q}{q}$ ( $a, b \ge 0$ ) completely, using the concavity of $\log$ .

ab \le \frac{a^p}{p} + \frac{b^q}{q}, \qquad \text{equality} \iff a^p = b^q

p = 2

,

n = 8

: Young's inequality feeds each term

a_k b_k \le \frac{a_k^2}{2} + \frac{b_k^2}{2}

, and summing gives Hölder (Cauchy–Schwarz).

Show proof

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 Jensen gives equality only when $X = Y$ , i.e. $a^p = b^q$ . This equality condition becomes Hölder's equality case in problem 3 below.

Derive Hölder's inequality $\sum_{k=1}^{n} |a_k b_k| \le \left(\sum |a_k|^p\right)^{1/p}\left(\sum |b_k|^q\right)^{1/q}$ completely from Young's inequality by the normalization argument.

\sum A_k B_k \le \sum\left(\frac{A_k^p}{p} + \frac{B_k^q}{q}\right) = \frac{1}{p} + \frac{1}{q} = 1

p = 3

(

q = 1.5

),

n = 8

: left side

H_8 \approx 2.7179

, right side

\approx 4.245

— the gap is positive and the inequality holds.

Show proof

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.

Prove that equality in Hölder's inequality holds exactly when $|a_k|^p$ is proportional to $|b_k|^q$ : $|a_k|^p = C\,|b_k|^q$ for all $k$ and some constant $C \ge 0$ .

\text{equality in Hölder} \iff A_k^p = B_k^q \;\forall k \iff |a_k|^p = C\,|b_k|^q

p = 2

,

n = 8

:

a_k = 1/k

is not proportional to the constant sequence

b_k = 1

, so the gap is nonzero — showing why the equality condition fails.

Show proof

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 (its sum $\frac{1}{p}\sum A_k^p + \frac{1}{q}\sum B_k^q = 1$ is an identity). So equality overall holds if and only if Young's inequality is an equality for every $k$ .
By problem 1's equality condition, this is equivalent to $A_k^p = B_k^q$ for every $k$ . Unwinding $A_k = |a_k|/\|a\|_p$ , $B_k = |b_k|/\|b\|_q$ 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 $C = \|a\|_p^p/\|b\|_q^q \ge 0$ independent of $k$ .
The converse holds too: $|a_k|^p = C|b_k|^q$ gives $A_k^p = B_k^q$ after normalization, making every Young's inequality an equality. In the scene, $a_k = 1/k$ is not constant, so with $b_k = 1$ the gap can never be 0 — the gap closing as $p \to 1^+$ is a collapse of the right side itself, not proportionality.

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