Why is the unit sphere strictly convex w.r.t "continuous" combinations? Let $\mu$ be a probability measure $\mathbb{R}^n$ (on the Borel or Lebesgue $\sigma$-algebra, I do not really care) which is supported on $\mathbb{S}^{n-1}$. (You can just think on a measure on the sphere itself).
Suppose that for some unit vector $v \in \mathbb{S}^{n-1}$, it holds that
$$ v= \int_{\mathbb{S}^{n-1}} x \, d\mu(x)$$
(That is $v$ is a "continuous" convex combination of the points of the sphere).
How to prove that $\mu = \delta_v$?
(I know this holds for finite combinations, and I am not sure how to generalize to the continuous case)
 A: Suppose $\mu$ is a Borel probability measure supported on $S=S^{n-1}$ satisfying the stated equality. We want to show that $\mu=\delta_\nu$ which is equivalent to $\mu(S\setminus\{\nu\})=0$. Now, $$S\setminus \{\nu\} = \bigcup_{k\geq 1} \left(S \setminus B\left(\nu,\frac1{n}\right)\right)$$ is a countable union of compact sets. So by $\sigma$-additivity  it suffices to show that $\mu(K)=0$ for every compact set $K\subset S^{n-1}\setminus\{\nu\}$.  So suppose this is not the case and that  $r:=\mu(K)>0$ for some such $K$. 
Since $K$ is compact and disjoint from $\nu$: $$\sup_{x\in K}\nu\cdot x = \max_{x\in K}\nu\cdot x=:k<1$$
And then
 $$1=\nu\cdot \nu =\int_S \nu\cdot x d\mu = \int_K \nu\cdot x d\mu + \int_{S-K}  \nu\cdot x d\mu \leq k r + (1-r) 1 <1 $$ Contradiction.
A: Here is an answer based on the fact that if you take a subset $A\subset \mathbb S^{n-1}$ of positive measure then $\frac{1}{\mu(A)}\int_A xd\mu(x)$ lies in the closed convex hull of $A$ $(*)$. One way to prove that is to use the idea of Ethan Bolker and remember you can always chose $\mu_n=\sum_i \lambda_i^n \delta_{x_i^n}$ such that the $\lambda$ are $\geq 0$ and such that $\mu_n$ is a probability measure. I prove that result in the "bonus section".
Take a look at $\mu(B(\nu,1/k)\cap\mathbb S^{n-1})$ as $k$ goes to infinity and $B(\nu,r)$ is the open ball of diameter $2r$ centered in $\nu$.
If $\mu(B(\nu,1/k)\cap\mathbb S^{n-1})$ is always equal to one then $\mu=\delta_\nu$ and we're done. Othewise there is some $k$ such that $\mu(B(\nu,1/k)\cap\mathbb S^{n-1})=\alpha<1$. Now $\displaystyle\frac{1}{\alpha}\int_{B(\nu,1/k)\cap\mathbb S^{n-1}}xd\mu(x)$ is of norm at most one and in the closed convex hull of $B(\nu,1/k)\cap\mathbb S^{n-1}$ while $\displaystyle\frac{1}{1-\alpha} \int_{S^{n-1}\backslash B(\nu,1/k)}xd\mu(x)$ is of norm at most one and is a point in the closed convex hull of $S^{n-1}\backslash B(\nu,1/k)$, which does not countain $\nu$. Since the sphere is the set of extremal points of the euclidian ball we conclude that $$\int_{B(\nu,1/k)\cap\mathbb S^{n-1}}xd\mu(x)+\int_{S^{n-1}\backslash B(\nu,1/k)}xd\mu(x),$$ which is the barycenter of the two previous integrals with mass $\alpha$ and $1-\alpha$, is not on $\mathbb S^{n-1}$. Thus if $\mu\neq\delta_\nu$ then $\int_{\mathbb S^{n-1}}xd\mu(x)\neq \nu$. This conclude the proof. Note that the proof is not so hard once we have proved $(*)$, and the property $(*)$ is merely saying that taking the integral of the identity function with respect to a probability measure is indeed a continuous convexe combination.
Remark : $x$ is called an extremal point of $C$ if and only if for $t\in (0;1)$ and $y,z\in C$ one have $ty+(1-t)z=x \Rightarrow y=z$.
BONUS : Since there are few questions about this weak star convergence i'll elaborate a bit. Let $f$ be a continuous function on the sphere and let $\varepsilon$ be some positive real. Take $\eta>0$ such that $|x-y|\leq \eta \Rightarrow |f(x)-f(y)|\leq \varepsilon$. Take $U_1,\ldots,U_N$ disjoint subsets of the sphere, of diameter smaller than $\eta$ and such that $\mathbb S^{n-1}=\bigcup U_i$. Moreover take $x_1,\ldots,x_N$ points of the sphere such that $x_i\in U_i$. Now the measure $\mu_\eta=\sum \mu(U_i)\delta_{x_i}$ is a probability measure and all the $\mu(U_i)$ are $\geq 0$ and it's close to $\mu$ in the weak star topology. Indeed you have $|f(y)-f(x_i)|\leq \varepsilon $ for every $y\in U_i$ so $$\left|\int_{U_i}fd\mu-\int_{U_i}fd\mu_\eta\right|=\left|\int_{U_i}fd\mu-\mu(U_i)f(x_i)\right|=\left|\int_{U_i}f-f(x_i)d\mu\right|\leq \int_{U_i}|f(y)-f(x_i)|d\mu(y)$$
$$\leq \varepsilon \mu(U_i)$$
Summing this over every $i$ we get $$\left|\int_{\mathbb S^{n-1}}fd\mu-\int_{\mathbb S^{n-1}}fd\mu_\eta\right|\leq \sum_i \left|\int_{U_i}fd\mu-\int_{U_i}fd\mu_\eta\right|\leq \sum_i \varepsilon \mu(U_i)\leq \varepsilon .$$
So we have indeed the convergence of the $\mu_eta$ to $\mu$ in the weak star topology. 
Now take $A$ a subset of the sphere with positive measure. $\frac{1}{\mu(A)}\mu_{|A}$ is a probability measure and is the weak star limit of the probability measures $\sigma_n=\sum_i \lambda_i^n \delta_{x_i^n}$ (with $\lambda_i^n$ positive, adding to one and the $x_i$ in $A$). So we have 
$$\frac{1}{\mu(A)}\int_A x d\mu(x)=\lim_n \int_A xd\sigma_n(x)=\lim_n \sum_i \lambda_i^n x_i^n.$$
Since $\sum_i \lambda_i^n x_i^n$ is in the convex hull of $A$ for every $n$ the limit is in the closed convex hull of $A$.
A: Too long for a comment, not an answer.
I haven't thought about this kind of problem in many years. That said, you may be able to show that $\mu$ is the (weak-star?) limit of measures supported by finitely many points. And that might lead to a proof.
Reference:
A Class of Convex Bodies
Ethan D. Bolker
Transactions of the American Mathematical Society
Vol. 145 (Nov., 1969), pp. 323-345 
https://www.jstor.org/stable/1995073?seq=1#page_scan_tab_contents
A: This is an attempt to carry through the approahc suggested by Ethan:
Indeed, it is true that the subspace of measures supported by finitely many points is dense in the weak-star topology of $M(\mathbb{S}^{n-1})$ (See here). However, I am not sure whether every probability measure $\mu$ admits a sequence of finitely supported probability measures $\mu_n$.
Let $\mu_n \to \mu$. Fix $1 \le i \le n$. Then $\int_{\mathbb{S}^{d-1}} x_i \, d\mu_n(x) \to \int_{\mathbb{S}^{d-1}} x_i \, d\mu(x)$, so
$$(*) \, \, \int_{\mathbb{S}^{d-1}} x \, d\mu_n(x) \to \int_{\mathbb{S}^{d-1}} x \, d\mu(x)=v$$
Suppose $\mu_n$ is supporetd on the set $\{x_1^n,x_2^n,...x_{r_n}^n\}$, i.e $$\mu_n=\sum_{i=1}^{r_n} \lambda_i^n\delta_{x_i^n},$$
then by $(*)$ we get
$$\lim_{n \to \infty} \sum_{i=1}^{r_n} \lambda_i^n x_i^n=v$$
It is not clear how to continue from here. Ideally, we would like to show that either $\lambda_i^n \to 0$ or the $x_i^n$ converges to $v$. (that is, the combinations tend to the trivial combination).
Part of the problem seems to be that the $\lambda_i^n$ are not known in advance to be non-negative (i.e we are taking here about arbitrary linear ("akmost-affine"**) combintations, not convex ones)

**The combinations are almost affine in the sense that
$$\sum_{i=1}^{r_n} \lambda_i^n =\mu_n(\mathbb{S}^{d-1}) \to  \mu(\mathbb{S}^{d-1})=1$$
