# Hilbert space $H$ is strictly smooth

I am trying to show that every Hilbert space $H$ is strictly smooth with modulus of smoothness $\phi_H(t)=\sqrt{1+t^2} -1$.

To show this I think I should show $H$ is uniformly smooth first. $\|\frac{h+tg}{2}\|+\|\frac{h-tg}{2}\|=1+o(t)$ uniformly for $\|h\|=1$, $\|g\|\le 1$. Modulus of smoothness is $\phi_H(t)= \sup_{\|h\|=1, \|g\|\le 1} \|\frac{h+tg}{2}\|+\|\frac{h-tg}{2}\| -1$ And then I don't know how to show $H$ is strictly smooth.

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And what does "strictly smooth" mean? – Martin Argerami Apr 3 '12 at 5:34

Start with $$\|x+ty\| + \|x-ty\| = \sqrt{(\|x+ty\| + \|x-ty\|)^2} \leq \sqrt{2(\|x+ty\|^2 + \|x-ty\|^2)}$$ and use the parallelogram identity to get (with $\|x\|=\|y\|=1$) $$\|x+ty\| + \|x-ty\|\leq \sqrt{4(\|x\|^2+t^2\|y\|^2)} = 2\sqrt{1+t^2}.$$ Then you are almost done.