Why is the induced Hilbert space norm $\|\cdot\|^2$ strictly convex?

Question

Let $(H, \langle \cdot,\cdot \rangle)$ be a Hilbert space with the induced norm defined by $v \mapsto \|v \| := \big( \langle v, v \rangle \big)^{1/2}$.

How can we show that $\| \cdot \|^2$ is strictly convex?

Answer 1

Let $v, w \in H$ with $v \neq w$ and $\alpha \in (0,1)$. \begin{align*} &\alpha \|v\|^2 + (1-\alpha) \|w\|^2 - \|\alpha v + (1-\alpha)w\|^2 \\ = &\alpha \|v\|^2 + (1-\alpha) \|w\|^2 - \big( \alpha^2 \|v\|^2 + (1-\alpha)^2 \|w\|^2 + 2\alpha(1-\alpha) \langle v,w \rangle \big) \\ = &(\alpha - \alpha^2) \|v\|^2 + \underbrace{\big((1-\alpha) - (1-\alpha)^2\big)}_{= \alpha - \alpha^2} \|w\|^2 - 2 \alpha(1-\alpha)\langle v,w \rangle \\ = &\alpha(1-\alpha) \big( \|v\|^2 + \|w\|^2 - 2\langle v,w \rangle \big) \\ = &\alpha(1-\alpha) \underbrace{\|v - w\|^2}_{\substack{\neq 0 \\ \textrm{since $v \neq w$}}} >0 \end{align*} That is $$ \|\alpha v + (1-\alpha)w\|^2 < \alpha \|v\|^2 + (1 - \alpha) \|w\|^2, $$ so $\|\cdot\|^2$ is strictly convex.

Note: We didn't require completeness, so this holds for any inner product space.