Given that $x \in \Bbb{R}^n$ and $p, q \in (1, \infty)$, maximize $||x||_q$ subject to $||x||_p = 1$

Question

How do I maximize $||x||_q$ subject to $||x||_p = 1$? Are there any useful inequalities I could consider? I know that if $0 < p < q$, then $||x||_p \ge ||x||_q$, so $||x||_q \le 1$. This is attainable with $x = (1, 0, 0, ..., 0)$ so that is the maximum.

If $p = q$, then it is obviously just $1$.

The other case is $q < p$ in which case $||x||_q \ge ||x||_p = 1$. But, well, I don't know how to show it must be $\le 1$ as well. Perhaps there are some other inequalities I could use to obtain an upper bound?

Is there a different way to do it? We haven't been taught Lagrange Multipliers yet — John, Oct 04 '20 at 21:04

score 2 · Accepted Answer · edited Oct 05 '20 at 11:00

Using what proved here we know that $$\left \| x \right \|_q \leq n^{\frac{1}{q}-\frac{1}{p}}\left \| x \right \|_p$$ Which means that if $\left \| x \right \|_p=1$ we can get at most the value $n^{\frac{1}{q}-\frac{1}{p}}$. However, this is also the maximum itself since for $x_{opt}=n^{-\frac{1}{p}}\mathbf{1}$ of $p$-norm $1$ $$\left \| x_{\text{opt}} \right \|_q =\left ( n\cdot n^{-\frac{q}{p}} \right )^{\frac{1}{q}}=n^{\frac{1}{q}-\frac{1}{p}}$$

Given that $x \in \Bbb{R}^n$ and $p, q \in (1, \infty)$, maximize $||x||_q$ subject to $||x||_p = 1$

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