# Equality case in elementary form of Holder's Inequality

A well known elementary formulation of Holder's Inequality can be stated as follows:

Let $a_{ij}$ for $i = 1, 2, \dots, k; j = 1, 2, \dots, n$ be positive real numbers, and let $p_1, p_2, \dots, p_k$ be positive real numbers such that $\sum_{i=1}^{k} \frac{1}{p_i} = 1$. Then we have

$$\sum_{j = 1}^{n} \prod_{i = 1}^{k} a_{ij} \leq \prod_{i=1}^{k} \left ( \sum_{j = 1}^{n} a_{ij}^{p_i} \right )^{\frac{1}{p_i}}.$$

Does anyone know when equality occurs in this inequality? Any insights on the equality case and/or a proof would be appreciated.

I have seen several proofs of the two sequence case. Here is one, from Cvetkovski's Inequalities:

Perhaps this argument can be generalized to prove the above result. Thanks in advance!

## 1 Answer

Since the usual proof of generalized Holder's proceeds by induction from the usual Holder's inequality, the equality case is precisely the same. In other words, if $v_j$ denotes the vector with components $a_{ij}^p$, then equality occurs iff all the vectors $v_j$ are parallel.

• Thank you -- could you direct me to a source where I can see the details of the proof? – WeierstrassSauce Jan 3 '15 at 19:55
• I think most first-year real analysis courses cover a proof of the generalized integral version. It was an exercise in our book (Folland's Real Analysis). Someone wrote up a solution to that exercise here: math.rochester.edu/people/grads/clungstr/real_analysis/… – pre-kidney Jan 3 '15 at 20:00