Let $1 \leq p < q \leq \infty$ ($p$ and $q$ are not otherwise related).

Given $\|x\|_q\leq\|x\|_p $ $\forall$ $ x \in \mathbb R^n$ how can I use Hölder's inequality to show $\|x\|_p\leq n^{\frac{1}{p}-\frac{1}{q}}\|x\|_q$ .

I can see this link is a related topic but I could not recognize Hölder's inequality usage in there. Am I missing anything?

I am assuming I need to say $\|x\|_p \leq \|x\|_1\leq...$ and here I need to find functions $f$ $g$ such that $\|x\|_1 \leq \|fg\|_1$. Is there a general "rule of thumb" for picking functions on these cases?

  • 1
    $\begingroup$ In addition to the answer below, another easy (but stupid) way to solve this problem is to use the Riesz-Thorin theorem. Indeed, just consider the identity map from $\Bbb R^n$ to itself, and interpolate between $p$ and $\infty$ (with respect to the $p$-norm in the codomain). Note that the operator bound on $L^p$ is $1$ and the operator bound on $L^{\infty}$ is $n^{1/p}$, so by Riesz-Thorin you exactly get $n^{1/p-1/q}$ as the operator bound on $L^q$. $\endgroup$ – shalop May 7 '16 at 23:29

The trick is to use the constant function $1$.

Let $r$ be the Hölder dual exponent of $q/p$ (which is $\ge 1$ by the assumption), that is $\frac1{q/p} + \frac1r=1$. Then we have:

$$ \|x\|^p_p = 1\cdot |x_1|^p + \cdots + 1\cdot |x_n|^p \le \left(1^r+\cdots+1^r\right)^{1/r} (|x_1|^q+\cdots+|x_n|^q)^{p/q}=n^{1/r} \|x\|_q^p$$

So $\|x\|_p \le n^{\frac1p-\frac1q} \|x\|_q$.

As you can see this is independent of the inequality $\|x\|_q\le \|x\|_p$ the proof of which does not require Hölder's inequality (you just normalize and use that $|t|^q\le |t|^p$ for $0\le t\le 1$, as shown in answer you linked). So you didn't miss anything; there is no use of Hölder's inequality in that linked answer.

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