Show that the unit sphere is strictly convex I can prove with the triangle inequality that the unit sphere in $R^n$ is convex, but how to show that it is strictly convex?
 A: I will show that if $(E,\langle\dot,\dot\rangle)$ is an inner product space, then the normed space $(E,\|\dot\|)$ is strictly convex, where $\|x\|:=\sqrt{\langle x,x\rangle}.$
Take any two point $x,y\in E,$ with $\|x\|=\|y\|=1$ and $x\neq y.$
Then for any $0<\alpha<1,$ we have $\|\alpha x+(1-\alpha) y\|<1,$ or equivalently $\|\alpha x+(1-\alpha) y\|^2-1<0.$ Infact:
$$\|\alpha x+(1-\alpha) y\|^2-1=\alpha^2+(1-\alpha)^2+2\alpha(1-\alpha)\langle x,y\rangle-1=2(1-\langle x,y\rangle)(\alpha-1)\alpha$$
Which is negative for $\alpha\in]0,1[$ because the Cauchy inequality and the hypothesis $\|x\|=\|y|=1,$ $x\neq y,$ imply $-1\leq\langle x,y\rangle<1.$
A: For the euclidean metric, we see that the unit ball  is strictly convex because for different vectors $a$ and $b$ we have that
$$f(t):=||ta+(1-t)b||^2\\ =\langle ta+(1-t)b,  ta+(1-t)b\rangle\\ = t^2||a||^2+(1-t)^2||b||^2 + 2t(1-t)\langle a,b\rangle\\= (\ldots) t^2+(\ldots)t + (\ldots)$$
is a quadratic function with positive leading term (because $f(t)\to+\infty$ as $t\to\infty$), hence $f$ assumes its maximum on the interval $[0,1]$ only at one or both of the endpoints.
In other words: We make use of the fact that  $f(t)=c_2t^2+c_1t+c_0$ with $c_2>0$ is strictly convex, which implies $f(t)<\max\{f(0),f(1)\}$ for $0<t<1$.
A: To show that the closed unit ball $B$ is strictly convex we need to show that for any two points $x$ and $y$ in the boundary of $B$, the chord joining $x$ to $y$ meets the boundary only at the points $x$ and $y$.
Let $x,y \in \partial B$, then $||x|| = ||y|| = 1.$ Now consider the chord joining $x$ to $y$. We can parametrise this by $c(t) := (1-t)x + ty$. Notice that $c(0) = x$ and $c(1) = y$. We need to show that $c(t)$ only meets the boundary when $t=0$ or $t=1$. Well:
$$||c(t)||^2 = \langle c(t), c(t) \rangle = (1-t)^2\langle x, x \rangle + 2(1-t)t \, \langle x,y \rangle + t^2 \langle y,y \rangle$$
$$||c(t)||^2 = (1-t)^2||x||^2 + 2t(1-t)\langle x,y \rangle + t^2||y||^2$$
Since $x,y \in \partial B$ it follows that $||x|| = ||y|| = 1$ and so:
$$||c(t)||^2 = (1-t)^2 + 2t(1-t)\langle x,y \rangle + t^2 \, . $$
If $c(t)$ meets the boundary then $||c(t) || = 1$, so let's find the values of $t$ for which $||c(t)|| = 1$:
$$(1-t)^2 + 2t(1-t)\langle x,y \rangle + t^2 = 1 \iff 2t(1-t)(1-\langle x, y \rangle) = 0 \, .$$
Clearly $t=0$ and $t=1$ are solution since $c(0)$ and $c(1)$ lie on the boundary. Recall that $\langle x, y \rangle = \cos\theta$, where $\theta$ is the angle between vectors $\vec{0x}$ and $\vec{0y}$, because $||x|| = ||y|| = 1.$ Thus, provided $x \neq y$ we have $\langle x, y \rangle \neq 1$ and so the chord only meets the boundary at $c(0)$ and $c(1).$
