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Why the following norm of $C[0,1]$ is strictly convex

$||| f|||= ||f||_\infty + ||f||_{L^2 [0,1]}$

where $||f||_{L^2 [0,1]}$ refers to $p$-norm of $L^p[0,1]$ when $p=2$.

I'm not sure if this relevant but I did prove that $||f||_\infty$ isn't strictly convex and $||f||_{L^2 [0,1]}$ is strictly convex.

Thanks

( A norm || || is strictly convex if when ||x||=||y||=1 and ||x+y||=2 so x=y )

user18217
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  • If one of the norms is stricly convex then their sum is stricly convex as well, isn't it? – Rasmus Nov 08 '11 at 17:35
  • the problem is that strictly convexity deals with vectors that have the same norm, but in that case the fact that ||f||(L^2 [0,1])= ||g||(L^2 [0,1]) doesn't mean that |||f|||=|||g||| – user18217 Nov 08 '11 at 17:39
  • Just to make sure we are talking about the same thing, could you add the definition of a strictly convex norm to your question? – Rasmus Nov 08 '11 at 20:34
  • In convex analysis/optimization, strict convexity means $x \ne y, \theta \in (0,1)$ implies $f(\theta x + (1-\theta)y) < \theta f(x) + (1-\theta)f(y)$, so norms aren't strictly convex, due to homogeneity. Rasmus may have been thinking of that definition. – p.s. Nov 08 '11 at 21:54

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Hint: if $f$ is not (a.e.) a scalar multiple of $g$, $|(f,g)| < \|f\|_2 \|g\|_2$ (where $(f,g)$ is the $L^2$ inner product), so $\|f+g\|_2 < \|f\|_2 + \|g\|_2$.

Robert Israel
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