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I've seen in lecture notes that norm on any Hilbert space is strictly convex means

"$\|x\|=\|y\|=1, \quad\|x+y\|=2 \Rightarrow x=y$"

But why this means strict convexity? I thought strict convexity means $\forall x_{1} \neq x_{2} \in H, \forall t \in(0,1) : \quad \|t x_{1}+(1-t) x_{2}\|<t \|f\left(x_{1}\right)\|+(1-t)\| f\left(x_{2}\right)\|$?

It would be best if you can please give a complete proof of why norm on "any" Hilbert space is strictly convex

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  • $\begingroup$ Perhaps it would help you to think of strict convexity in terms of an inequality: if $x \neq y$ are two points for which $||x||=1=||y||$, it follows that $$||(x+y)/2||<1$$ $\endgroup$ – TM Gallagher Apr 8 at 1:50
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    $\begingroup$ My answer here may help. $\endgroup$ – Theo Bendit Apr 8 at 2:14
  • $\begingroup$ Possible duplicate: math.stackexchange.com/questions/2133559/… $\endgroup$ – avs Apr 8 at 5:21
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    $\begingroup$ @avs A helpful link, but not really a duplicate. The asker is more confused about how the (seemingly) weaker form of strict convexity implies the stronger form. $\endgroup$ – Theo Bendit Apr 8 at 10:08

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