# uniform convergence of continuous functions

Let $X=C[0,1]$ and $x,x_{n}\in X,\forall n\in\mathbb{N}.$ Suppose $\forall t\in[0,1],\ x_{n}(t)\rightarrow x(t)$ and $\sup_{n}\left\Vert x_{n}\right\Vert <\infty$. Show that there exist convex combination of $x_{n}$ that converge to $x$ in $X$ uniformly.

It seems to be closed related to the following theorem:

Suppose $X$ is a metrizable locally convex space. If $\{x_{n}\}$ is a sequence in $X$ that converges weakly to some $x\in X$, then $\exists$ a convex combination of $x_{n}$ that converges to $x$ originally.

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Take $\mu$ a (positive) finite Radon measure on $[0,1]$, and use dominated convergence to show that $\int_{[0,1]} x_nd\mu \to \int_{[0,1]} xd\mu$. Conclude by Riesz representation since any functional on $C[0,1]$ is given by integration with respect to the difference of two such measures. – Jose27 Apr 30 '12 at 2:48
Riesz representation theorem seems to apply to linear functionals. But C[0,1] meas all continuous functions on [0,1]. Am I wrong? – user16859 Apr 30 '12 at 3:21
Linear functionals on $C[0,1]$ are given by derivatives of functions of bounded variation which define Radon measures. – Jose27 Apr 30 '12 at 3:44
Yes, sure, but any vector space is convex by definition so... – Jose27 Apr 30 '12 at 5:16
@Jose27 Why not writing your first comment as an answer? – martini Apr 30 '12 at 6:07

Okay, so the point is to prove weak convergence, then the theorem you mention would automatically give you norm convergence, ie. uniform convergence (hopefully this answers your last question). To do this we notice that since to every functional $l\in C[0,1]^*$ we can associate two Radon measures such that $l(f)=\int fd\mu_1-\int fd\mu_2$, so it's enough to show that $\int x_n d\mu \to \int xd\mu$, but this is just dominated convergence since $\mu$ is finite and $\| x_n\| \leq C$ for some $C$ which does not depend on $n$.