Let $1\le p\le q$ and $x\in \mathbb{R}^n$. Show that $$ n^{\frac{1}{q}}\|x\|_p \le n^{\frac{1}{p}} \|x\|_q, $$ where $\|x\|_k = (\sum_{i}|x_i|^k)^\frac{1}{k}$.
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There is a similar question, but I can't use the same technique to prove this inequality. – Chao Liao Sep 15 '15 at 07:06
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I think you could reason similarly, using the sup metric to get an intermediate inequality – Anthony Peter Sep 15 '15 at 07:16
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That's a standard application of Hölder's inequality. Let $\alpha := \frac qp \ge 1$ and $\beta = \frac 1{1 - \frac 1\alpha} = \frac q{q-p}$ \begin{align*} \def\norm#1{\left\|#1\right\|}\def\abs#1{\left|#1\right|} \norm{x}_p &= \left(\sum_i \abs{x_i}^p\right)^{1/p}\\ &= \left(\sum_i \abs{x_i}^p \cdot 1 \right)^{1/p}\\ &\le \left(\norm{(\abs{x_i}^p)_i}_\alpha \norm{1}_\beta\right)^{1/p}\\ &= \left(\sum_i \abs{x_i}^{p\alpha}\right)^{1/p\alpha} \cdot \left(\sum_i 1^{\beta}\right)^{1/p\beta}\\ &= \norm x_q \cdot n^{(q-p)/pq}\\ &= \norm x_q \cdot n^{1/p - 1/q} \end{align*}
martini
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Excellent answer. Thank you! But there is small mistake that $\alpha$ should be $\frac{q}{p}$. – Chao Liao Sep 15 '15 at 07:59
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Here is a proof using Jensen's inequality. Note that $\phi(x) = x^{\tfrac q p}$ is convex since $q \ge p$. Now we can apply Jensen's ineq $\phi(E X) \le E\phi(X)$. Our distribution is uniform over the set $\{\lvert x_i\rvert^p \mid i=1,\dots, n\}$.
$$\begin{align} & \left[ \frac{1}{n}\sum_{i=1}^{n} \lvert x_i\rvert^p \right]^{q/p} &\le \frac{1}{n}\sum_{i=1}^{n} \lvert x_i\rvert^q \\ & \implies \frac{1}{n^{1/p}} \lVert x\rVert_p &\le \frac{1}{n^{1/q}} \lVert x\rVert_q \end{align}$$
seeker
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