Uniform convergence of Lagrange polynomials There is a well-known theorem that states that on a closed interval $[a,b]$ any continuous function is the limit of a uniformly convergent sequence of polynomials.
Proofs for this theorem usually resort to Bernstein polynomials.
What about if we try with Lagrange polynomials ?
WLOG we may assume $[a,b]=[0,1]$
For a given $n \in \mathbb N$, let $x_1,...,x_n \in [0,1]$ be arbitrary and consider $$L_n(x)=\sum_{k=1}^n f(x_k)\prod_{i=1,i\neq k}\frac{x-x_i}{x_k-x_i}$$ 
which is known as the interpolating Lagrange polynomial for points $\{(x_1,f(x_1)),...,(x_n,f(x_n)) \}$.


*

*Does the sequence of polynomials $(L_n)$ converge uniformly to $f$ ?

*If not, does choosing points smartly instead of randomly leads to uniform convergence ? 

 A: It depends on the function $f$ and I'll assume that the interpolation points are distinct. The closest result I know is the following: 
Theorem:
If $f:[0,1]\rightarrow\mathbb{C}$ is analytic and analytically continuable to a function that is analytic in a region containing the closed "stadium" of radius $1$ (consisting of all the points in the complex-plane that are at a distance $\leq1$ from $[0,1]$). Then, $L_n$ converges uniformly to $f$ as $n\rightarrow \infty$. The convergence is also geometrically fast. 
(The proof uses the Hermite integral formula for $f - L_n$.) The result can be found on page 82 of the book Approximation Theory and Approximation Practice (ATAP) by L. N. Trefethen. 
In particular, no assumption on the distribution of the interpolation points is made (only that they are distinct). They can all be on $[0,1/2]$ if you wish.
For functions that do not satisfy the above Theorem then I am not sure what can be said about the convergence of $L_n$ for general interpolation points.
For specific interpolation points, lots is known. For example: 


*

*Chebyshev points: If $f$ is Lipschitz continuous, then $L_n$ converges to $f$ uniformly (also in ATAP). I think $f$ is Holder continuous with exponent $>0$ will also do. 

*Leja points: If $f$ is analytic on $[0,1]$, then $L_n$ converges uniformly to $f$. 

*Equally-spaced points: Runge function is an example of an analytic function where the equally-spaced interpolants diverge from $f$ as $n\rightarrow\infty$.
A: Edit (rewritten):
If you assume $f$ is a smooth function, then one can show (see for example Theorem 4.3 here) that through Taylor series bounds:
$$f(x)=L_N(x)+e_N(x),$$
where,
$$e_N(x)=\frac{(x-x_1)(x-x_2)\cdots (x-x_N)f^{(N+1)}(c)}{(N+1)!}$$
for some fixed $c\in [0,1]$. So your error is bounded by $f^{(N+1)}(c)/(N+1)!$. 
