4
$\begingroup$

For any vector $x \in \mathbf{R}^{n}$, and any natural numbers $p \geq q \geq 1$, we have that $\lVert x \rVert_{p} \leq \lVert x \rVert_{q}$. Some proofs of this fact are given here and here. However, I am interested in proving this result using Holder's inequality. This is left as an exercise for the reader in An Introduction to Inequalities by Beckenbach and Bellman. In that text, they prove the case when $p = 3$ and $q = 2$ as follows. First, we write $$ \sum_{i=1}^{n} |x_{i}|^{3} = \sum_{i=1}^{n} |x_{i}| |x_{i}|^{2}. $$ We note that the right side of this equation is the inner product of $(|x_{1}| , \ldots , |x_{n}|)$ and $(|x_{1}|^{2} , \ldots , |x_{n}|^{2})$. Applying the Cauchy-Schwarz inequality to this inner product, we have that $$ \sum_{i=1}^{n} |x_{i}|^{3} \leq \left( \sum_{i=1}^{n} |x_{i}|^{2} \right)^{\frac{1}{2}} \left( \sum_{i=1}^{n} |x_{i}|^{4} \right)^{\frac{1}{2}}. $$ To continue, we note that $$ \left( \sum_{i=1}^{n} |x_{i}|^{2} \right)^{2} - \left( \sum_{i=1}^{n} |x_{i}|^{4} \right) = \sum_{i=1}^{n} \sum_{\begin{subarray}{c} j=1 \\ j \neq i \end{subarray}}^{n} |x_{i}|^{2} |x_{j}|^{2} \geq 0 , $$ where the inequality follows from the fact that a sum of nonnegative quantities is nonnegative. Rearranging this inequality, and taking square roots, we have that $$ \left( \sum_{i=1}^{n} |x_{i}|^{4} \right)^{\frac{1}{2}} \leq \sum_{i=1}^{n} |x_{i}|^{2}. $$ Applying this result to one of our earlier inequalities, we have that $$ \sum_{i=1}^{n} |x_{i}|^{3} \leq \left( \sum_{i=1}^{n} |x_{i}|^{2} \right)^{\frac{3}{2}}. $$ Taking the cube root of both sides yields the desired result. The text says that "the proof for arbitrary rational $m \geq n \geq 1$ can be obtained by a corresponding application of Holder's inequality." In analogy with the proof given above, using Holder's inequality instead of Cauchy's inequality, we have that $$ \sum_{i=1}^{n} |x_{i}|^{p} = \sum_{i=1}^{n} |x_{i}| |x_{i}|^{p-1} \leq \left( \sum_{i=1}^{n} |x_{i}|^{q} \right)^{\frac{1}{q}} \left( \sum_{i=1}^{n} |x_{i}|^{\frac{q (p-1)}{q-1}} \right)^{\frac{q-1}{q}} \leq \left( \sum_{i=1}^{n} |x_{i}|^{q} \right)^{\frac{p}{q}}. $$ In order to justify this chain of inequalities, we still need to show that $$ \left( \sum_{i=1}^{n} x_{i}^{\frac{q(p-1)}{q-1}} \right)^{\frac{q-1}{q}} \leq \left( \sum_{i=1}^{n} x_{i}^{q} \right)^{\frac{p-1}{q}}. $$ Letting $z_{i} = x_{i}^{\frac{q}{q-1}}$, this inequality is equivalent to $$ \left( \sum_{i=1}^{n} z_{i}^{p-1} \right)^{q-1} \leq \left( \sum_{i=1}^{n} z_{i}^{q-1} \right)^{p-1}. $$ But I am stuck at this point. (Recall that $p \geq q \geq 1$.) Any thoughts? Note that this comes from an elementary textbook, so ideally the answer should only use precalculus mathematics.

$\endgroup$
3
$\begingroup$

Apply Hölder's inequality in the following way:

$$\sum_{i=1}^n |x_i|^p = \sum_{i=1}^n |x_i|^{q-1} |x_i|^{p+1-q} \le \left( \sum_{i=1}^n \left(|x_i|^{q-1}\right)^{\frac{q}{q-1}} \right)^{\frac{q-1}{q}} \left( \sum_{i=1}^n \left(|x_i|^{p+1-q}\right)^{q} \right)^{\frac{1}{q}} \le \left( \sum_{i=1}^n |x_i|^{q} \right)^{\frac{q-1}{q}} \left( \sum_{i=1}^n |x_i|^{q} \right)^{\frac{p+1-q}{q}} = \left( \sum_{i=1}^n |x_i|^{q} \right)^{\frac{p}{q}}.$$

$\endgroup$
  • $\begingroup$ Nice. I would only add that the second inequality is equivalent to $\sum |z_{i}|^{k} \leq (\sum |z_{i}|)^{k}$, which is obvious because the difference between the right and left sides of the inequality is the sum of all the cross terms in $(\sum |z_{i}|)^{k}$, all of which are nonnegative. $\endgroup$ – Stirling Sep 4 '13 at 18:13
  • $\begingroup$ I think the proof is incomplete. The second inequality is actually the special case that 1-norm is larger than any p-norm. The power is not integer, so you cannot use combinatorics to justify the inequality. $\endgroup$ – Tony Sep 29 '16 at 17:32
0
$\begingroup$

The last inequality you have to show is actually Hölder's inequality, but for the exponents $p':=p-1$ and $q':=q-1$. This suggests a proof by induction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.