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How would you prove that in a strictly convex normed vector space, the function $f(x) = \| x \|^2$ is strictly convex??


$E$ is strictly convex iff $\| t x + (1-t) y \| <1$ for all $x,y \in E$ such that $x \neq y$ and $||x||=||y||=1$.

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up vote 0 down vote accepted

I discovered that the key idea is to write

$$ \| \frac{t \|x\|}{t \|x\| + (1-t) \|y\|}\frac{x}{\|x\|} + \frac{(1-t) \|y\|}{t \|x\| + (1-t) \|y\|}\frac{y}{\|y\|} \|<1$$

and then the result follows easily.

Also, the result also holds for $f(x)=x^p$, with $p \geq 1$.

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Can you tell me more about your solution? – Irddo Mar 31 at 14:08

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