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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?

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Are you referring to the unit ball? –  Giuseppe Tortorella Nov 7 '12 at 17:38
    
@GiuseppeTortorella I'm talking about $||x||\leq 1$ –  André Nov 7 '12 at 17:39
    
Since there are norms for which the unit ball is not strictly convex (think 1-norm and max-norm), you should probably revise the question to specify which norm (e.g. Euclidean as in Hagen von Eitzen's Answer), or indicate that counterexamples are of possible interest. –  hardmath Nov 7 '12 at 17:56
    
@hardmath Thanks, I meant $R^n$. –  André Nov 7 '12 at 17:58
    
@André: But saying $\mathbb{R}^n$ only specifies the Cartesian set, not the norm involved in your Question. Wikipedia has more about strictly convex spaces. –  hardmath Nov 7 '12 at 18:10
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3 Answers

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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 $0<t<1$.

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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 $-1\leq\langle x,y\rangle<1.$

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+1 for pointing out the same proof given by @FlybyNight actually applies more generally, to any inner product space. –  hardmath Nov 8 '12 at 15:28
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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 $c(1).$

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