Hölder’s Inequality

Hölder’s inequality is Young’s inequality lifted from one product to a whole sum of products. Young controls a single mixed term. Hölder controls the total overlap of two sequences. The bridge is normalization.

The basic problem is this: a product like $𝑥_𝑛 𝑦_𝑛$ mixes two quantities together, and mixed terms are hard to control directly. Hölder’s inequality shows that if you measure each sequence in the right way first, then the total mixed overlap stays bounded.

Start with Young’s inequality: if \(p,q>1\) satisfy \[ \frac{1}{p}+\frac{1}{q}=1, \] then for any \(a,b\ge 0\), \[ ab\le \frac{a^p}{p}+\frac{b^q}{q} \] This says that a single product can be bounded by separate \(p\)- and \(q\)-power costs.

This is the local prototype for everything that follows. Hölder will not invent a new kind of control. It will repeat this same kind of control term-by-term and then add the results together.

Hölder’s inequality aims to prove that for sequences \(x=(x_n)\in \ell^p\) and \(y=(y_n)\in \ell^q\), \[ \sum_{n=1}^\infty |x_n y_n| \le \left(\sum_{n=1}^\infty |x_n|^p\right)^{1/p} \left(\sum_{n=1}^\infty |y_n|^q\right)^{1/q} \] The left-hand side is a sum of local products. The right-hand side is a product of global sizes. So the problem is to turn many local products into a bound in terms of global norms.

Here \(\ell^p\) denotes the space of all infinite sequences \(x=(x_n)\) such that \[ \sum_{n=1}^\infty |x_n|^p < \infty \] Its associated \(p\)-norm is \[ \|x\|_p=\left(\sum_{n=1}^\infty |x_n|^p\right)^{1/p} \] So saying \(x\in \ell^p\) means exactly that the sequence has finite \(p\)-norm Likewise, \(y\in \ell^q\) means \[ \sum_{n=1}^\infty |y_n|^q < \infty, \qquad \|y\|_q=\left(\sum_{n=1}^\infty |y_n|^q\right)^{1/q} \]

The key move is to normalize the sequences. Assume first that neither norm is zero, and define \[ a_n=\frac{|x_n|}{\left(\sum_{k=1}^\infty |x_k|^p\right)^{1/p}}, \qquad b_n=\frac{|y_n|}{\left(\sum_{k=1}^\infty |y_k|^q\right)^{1/q}} \] Then each sequence has unit size in its own scale: \[ \sum_{n=1}^\infty a_n^p = 1, \qquad \sum_{n=1}^\infty b_n^q = 1. \] Normalization strips away scale and leaves only shape.

Now apply Young’s inequality term-by-term: \[ a_n b_n \le \frac{a_n^p}{p}+\frac{b_n^q}{q} \] Summing over \(n\) gives \[ \sum_{n=1}^\infty a_n b_n \le \frac{1}{p}\sum_{n=1}^\infty a_n^p + \frac{1}{q}\sum_{n=1}^\infty b_n^q \] Since the normalized sums are both equal to \(1\), this becomes \[ \sum_{n=1}^\infty a_n b_n \le \frac{1}{p}+\frac{1}{q} =1. \] So once both sequences are normalized, their total overlap cannot exceed \(1\).

Now undo the normalization: \[ a_n b_n = \frac{|x_n|}{\|x\|_p}\frac{|y_n|}{\|y\|_q} = \frac{|x_n y_n|}{\|x\|_p\|y\|_q}, \] where \[ \|x\|_p=\left(\sum_{n=1}^\infty |x_n|^p\right)^{1/p}, \qquad \|y\|_q=\left(\sum_{n=1}^\infty |y_n|^q\right)^{1/q} \] Therefore \[ \sum_{n=1}^\infty \frac{|x_n y_n|}{\|x\|_p\|y\|_q}\le 1 \] Multiplying both sides by \(\|x\|_p\|y\|_q\) yields Hölder’s inequality: \[ \sum_{n=1}^\infty |x_n y_n| \le \|x\|_p\|y\|_q \] If one of the norms is zero, then the inequality is immediate.

The explanatory structure is: \[ \text{Young} \;\to\; \text{normalize} \;\to\; \text{apply term-by-term} \;\to\; \text{sum} \;\to\; \text{undo normalization} \;\to\; \text{Hölder} \]