Calculating Holder's Inequality for Sums with Exponents

Question

I'm a little confused about the procedure for calculating Holder's Inequalities for Sums with Exponents. For example, I tried to apply Holder's Inequality as follows

$$(\sum_{j=1}^{T}p_{j}^{(1/q) + (1/r) - 1})^{q} \leq (\sum_{j=1}^{T}p_j)(\sum_{j=1}^{T}p_j)^{q/r} $$

Is this valid?

It is hard to understand this unless you tell us what $q$ and $r$ are. It is false if $q>1$ and $r>1$. — Kavi Rama Murthy, Aug 28 '18 at 23:31
@KaviRamaMurthy I'm assuming q > 1, and that r satisfies 1/q + 1/r = 1 — Quinty, Aug 29 '18 at 00:07

score 0 · Accepted Answer · answered Aug 29 '18 at 00:10

Your inequality implies that if $p_i>0$ and $\sum p_i\leq 1$ then $\sum p_i^{t} \leq 1$ where $t=\frac 1 q +\frac 1 r -1$. This is easily seen to be false. Note that $t<1+1-1=1$. For example, take $p_i=\frac 1 T$ for $1\leq i \leq T$ and observe that the inequality fails for large values of $T$.

Calculating Holder's Inequality for Sums with Exponents

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