Given $x,y \in \mathbb{R}^n$, the Cauchy Schwarz inequality states,
$|x^Ty| \leq \|x\|_2\|y\|_2$
And for non-Euclidean (norms other than $l_2$), we have,
$|x^Ty| \leq \|x\|_p\|y\|_q$ where $\|\cdot\|_q$ is the dual norm of $\|\cdot\|_p$ (I remember $p,q$ is related by $1/p+1/q = 1$ or something like that)
For example,
$|x^Ty| \leq \|x\|_1\|y\|_\infty$
My question is:
Does Cauchy-Schwarz imply $|x^Ty| \leq \|x\|_p\|y\|_p$ for any $p
\geq 1$?
I feel like this should hold, based on the fact that all the norms are equivalent. In the previous example, we know that $\|y\|_\infty \leq \alpha \|y\|_1$ with some constant $\alpha$ (I can never remind the exact number)
Then it must be true that,
$|x^Ty| \leq \|x\|_1\|y\|_1$
Is there a flaw in my reasoning?
Best Answer
Well, actually we have $\|y\|_\infty \leq \|y\|_1$, and that implies $$ |x^Ty| \leq \|x\|_1\|y\|_\infty \le \|x\|_1\|y\|_1 \, , $$ so that your reasoning is essentially correct for $p=1$.
For $p > 1$ the same approach works as long as the conjugate $q$ (defined by $\frac 1p + \frac 1q = 1$ ) satisfies $p \le q$, that is, if $p \le 2$. Then Hölder's inequality and the relation between $p$-norms gives $$ |x^Ty| \leq \|x\|_p\|y\|_q \le \|x\|_p\|y\|_p \, . $$
The same reasoning does not work if $p > 2$, and in fact the desired estimate does not hold in that case. This can be seen by choosing $x = y \ne 0$: $$ |x^Ty| = \Vert x \Vert_2^2 > \Vert x \Vert_p^2 = \|x\|_p\|y\|_p \, . $$
Summary: $$ |x^Ty| \leq \Vert x\Vert_p\Vert y\Vert _p \text{ for all } x, y \in \Bbb R^n $$ holds if and only if $p \le 2$.