Let $f(x) = \exp(x)$ and $(\xi\_i)\_{i=0}^\infty, \, \xi\_i \in (0,1)$ be a sequence of points from the unit interval.


For $n \in \mathbb{N}$ let $P\_n$ be a polynomial of degree $n$ that interpolates $f$ at $\xi\_0, \ldots, \xi\_n$, i.e.
$$
f(\xi\_i)-P\_n(\xi\_i) = 0, \text{ for all } i=0,\ldots,n,
$$


**Question**:
Does uniform convergence of the interpolation process on $(0,1)$, i.e.
$$\lim\_{n \rightarrow \infty} \|f - P\_n\|\_\infty =0$$
 imply that the sequence $(\xi\_i)\_{i=0}^\infty$ is *dense* in $(0,1)$?
My intuition says it is but I failed to proof it.


Thank you!



Do interpolation nodes have to be dense?
Unfortunately not. Just look at the (confluent) limit case when all $\xi\_i=0$: in this case $P\_n$ is just the $n$-th Taylor polynomial for $f$. which clearly converges to $f$ not only on $(0,1)$, but on the whole complex plane, but $\xi\_i$ are all at the origin. 


A less extreme argument is: if we consider the sequence $q\_n$ of best polynomial approximants to $f$ on a subinterval $(\alpha, \beta)$ of $(0,1)$, then (i) $q\_n$ actually interpolate $f$ on $(\alpha, \beta)$ (but not outside), and (ii) $(f-q\_n) \to 0$ in the largest ellipse with foci at $\alpha$ and $\beta$ where $f$ has an analytic continuation. Since $f$ is entire, convergence holds on the whole $\mathbb{C}$, so on $(0,1)$.


EDIT after the comments of Dirk and Qiaochu Yuan:


I don't think that the fact that nodes are different or form a sequence makes any difference. Still, what about this argument: the Cauchy formula for the interpolation error at a point $t$ is
$$
(f-p\_n)(t)=\frac{f^{(n+1)}(\theta)}{(n+1)!} \omega\_n(t), \quad \omega\_n(t)=\prod\_{i=1}^{n+1}(t-\xi\_i).
$$
For $t\in (0,1)$, we have $\|\omega\_n\|\_\infty\leq 1$, and $f^{(n+1)}(\theta)\leq e$, REGARDLESS the distribution of $\xi\_i$'s on $(0,1)$. In particular, they could live on $(0, 1/2)$ only, etc.


