Why is the $L_p$ norm strictly convex for $1

Question

Let $x,y \in L_p$ such that $\|x\|_p=\|y\|_p=1$ , $1< p<\infty$ and $x\neq y.$ Why is $\|x+y\|_p<2$ ?

I'm not sure how to start the proof.. I don't know how to handle integral of $(x+y)^p$ and it seems that using the binomial theorem won't be a great success.

Start using the proof of the triangle inequality, and look at the equality case in Hölder inequality (you may have to use the strict convexity of the exponential). — Davide Giraudo, Nov 08 '11 at 10:28

score 15 · Accepted Answer · answered Nov 08 '11 at 10:27

15

Minkowski's Inequality gives $ \| x+y \|_p \leq \|x\|_p + \| y \|_p = 2 $ with equality if and only if $x$ and $y$ are positively linearly dependent (i.e. $x=ty$ or $y=tx$ for some $t\geq 0$).

If $x$ and $y$ were positively linearly dependent, say $x=ty$, then plugging into $\|x\|_p=\|y\|_p=1$ forces $t=1$ so $x=y$, which contradicts the assumption $ x\neq y.$ Thus we do not reach the equality case, so $\| x+y \|_p <2. $

answered Nov 08 '11 at 10:27

Ragib Zaman

35,127

Does this also hold in the complex case? – Lilla Feb 11 '21 at 19:59
@warm_fish Yes. – Ragib Zaman Feb 12 '21 at 02:23

Why is the $L_p$ norm strictly convex for $1

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