Is the map $x\mapsto\|x\|_p^p$, defined in $L^p[0,1]$ ($1

Question

Is the map $x\mapsto\|x\|_p^p$, defined in $L^p[0,1]$ (for $1<p<\infty$), strictly convex?

Maybe there is a classic inequality that can give the conclusion easily. Let me know if this is the case.

Edit: In the case it is not strictly convex, if we restrict the domain of $x\mapsto\|x\|_p^p$ to be the unit ball in $L^p[0,1]$, the map is strictly convex?

There is an answer to the equality case in Minkowski's inequality here: https://math.stackexchange.com/questions/581257/when-does-equality-hold-in-the-minkowskis-inequality-fg-p-leq-f-p-g — uniquesolution, Aug 17 '17 at 06:33

uniquesolution · Accepted Answer · 2017-08-17T07:43:55.203

1

We have for all $x,y$ and $\lambda\geq 0$, by the triangle (Minkowski) inequality: $$|\!|\lambda x+(1-\lambda) y)|\!|_p^p\leq\left(\lambda|\!|x|\!|_p+(1-\lambda)|\!|y|\!|_p\right)^p$$ and the result follows from the strict convexity of the function $t\to t^p$ for $p>1$ and $t\geq 0$.

edited Aug 17 '17 at 07:43

answered Aug 17 '17 at 06:44

uniquesolution

18,541

what if we take $x\neq y$ with $|x|_p=|y|_p$? – Filburt Aug 17 '17 at 07:36
@Filburt - Equality means equality also in the triangle inequality, which implies for $p>1$ that $x=cy$ for some $c\geq 0$, and if they also have the same norm then $c=1$. – uniquesolution Aug 17 '17 at 07:41

GuPe · Answer 2 · 2017-08-17T06:15:28.987

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Minkowski's inequality works: $$\begin{align}||\lambda g+(1-\lambda)f||_p=\left(\int_X (|\lambda g+(1-\lambda) f|)^p \text{d}\mu \right)^{\frac 1p} &\le \left(\int_X (\lambda |g|)^p \text{d}\mu \right)^{\frac 1p}+\left(\int_X ((1-\lambda) |f|)^p \text{d}\mu \right)^{\frac 1p}\\ &=\lambda ||g||_p+(1-\lambda)||f||_p \end{align}$$

edited Aug 17 '17 at 06:15

answered Aug 17 '17 at 06:13

GuPe

7,318

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hold on, how do we get the strictly inequality? ($\lambda\in]0,1[$) – Filburt Aug 17 '17 at 06:17
the equality holds if and only if $f$ and $g$ are linearly dependent... So if we suppose $f\neq g$ (different classes), is there a way to conclude $|\lambda g+(1-\lambda)f|_p<\lambda |g|_p+(1-\lambda)|f|_p$? thought they can be different and still linearly dependent – Filburt Aug 17 '17 at 06:20
@Filburt For $p=1$ equality can hold also for independent functions, provided they have non-intersecting supports. – uniquesolution Aug 17 '17 at 06:30
So I made another restriction ($1<p<\infty$) and edited the question – Filburt Aug 17 '17 at 06:32
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The question is asking about $|x|_p^p$, not $|x|_p$. – Did Aug 17 '17 at 06:32
@Did thats right – Filburt Aug 17 '17 at 06:33
$x^p$ is convex increasing, and composition of a convex function by a convex increasing function is convex – GuPe Aug 17 '17 at 06:37
@GuachoPerez "Strictly Convex" and "Convex" are not the same thing. – uniquesolution Aug 17 '17 at 06:38
But $x^p$ is strictly convex – GuPe Aug 17 '17 at 06:54

Is the map $x\mapsto\|x\|_p^p$, defined in $L^p[0,1]$ ($1

2 Answers2