I have been looking at an exercise about proving Hölder's inequality for $x, y \in \mathbb{R}^n$. I was wondering if you could help me to solve my confusion about the requirement that $p+q = pq$.
I think that given $p, q \geq 1$ the following derivation is correct:
$\lVert x \rVert_p \lVert y \rVert_q = \left(\sum_i \lvert x \rvert^p \right)^{1/p} \left( \sum_j \lvert y_j \rvert^q \right)^{1/q}= \left(\sum_i \lvert x \rvert^p \right)^{q/qp} \left( \sum_j \lvert y_j \rvert^q \right)^{p/qp} = \left(\sum_i \lvert x \rvert^{pq} \right)^{1/qp} \left( \sum_j \lvert y_j \rvert^{pq} \right)^{1/qp} = \left(\sum_i \lvert x \rvert^{pq} \cdot \sum_j \lvert y_j \rvert^{pq} \right)^{1/qp} = \left(\sum_{ij} \lvert x_i \rvert^{pq} \lvert y_j \rvert^{pq}\right)^{1/pq} = \left(\sum_{ij} \lvert x_i y_j \rvert^{pq} \right)^{1/pq} = \left(\sum_i \lvert x_i y_i \rvert^{pq} + \sum_{\substack{i,j \\ i\neq j}} \lvert x_i y_j \vert^{pq} \right)^{1/pq} \geq \left( \lvert \langle x, y \rangle \vert^{pq} +\sum_{\substack{i,j \\ i\neq j}} \lvert x_i y_j \vert^{pq} \right)^{1/pq} \geq \left(\lvert \langle x, y \rangle \rvert^{pq}\right)^{1/pq} = \lvert \langle x, y \rangle \rvert \geq \langle x, y \rangle$,
where $\langle .,. \rangle$ is the usual scalar product and all sums are up to $n$.
However, this doesn't require $p+q = pq$. Did I make a mistake somewhere or is the restriction on $p$ and $q$ superfluous?
Thank you for your help!