Does there exist $f \in \mathcal{C}^\infty\left([a, b]\right)$ such that $M_n \rightarrow \infty$ but $\left\|f - p_n\right\| \rightarrow 0$?

Question

Let $f \in \mathcal{C}^\infty\left([a, b]\right)$ and $n \in \mathbb{N}$.

Let $p_n$ be the Lagrange interpolating polynomial for a partition of equispaced points $a = x_0 < x_1 < \cdots < x_n = b$ through $f$, and $M_n = \left\|f^{(n)}\right\|_\infty$.

Firstly, I would show that if there exists $C > 0$ such that $M_n \leq C$ then $\lim\limits_{n \rightarrow \infty} \left\|f - p_n\right\|_\infty = 0$. I know that $\left\|f - p_n\right\|_\infty \leq \frac{M_{n + 1}}{(n + 1)!}(b - a)^{n + 1}$, so I think it's not difficult to prove it.

Nevertheless, does there exist $f \in \mathcal{C}^\infty\left([a, b]\right)$ such that $\lim\limits_{n \rightarrow \infty} M_n = \infty$ but $\lim\limits_{n \rightarrow \infty} \left\|f - p_n\right\| = 0$? I think it does exist, but which would be?

Answer 1

Consider $$f(x):=xe^x\quad\text{on}\quad[a,b]:=[0,1].$$

• $M_n=\max_{x\in[0,1]}\left(x+n\right)e^x=\left(n+1\right)e\to\infty$, but
• $\left\|f - p_n\right\|_\infty \le\frac{M_{n + 1}}{(n + 1)!}(b - a)^{n + 1}=\frac{\left(n+2\right)e}{(n + 1)!}\to0$.

More generally, on an arbitrary interval $[a,b]$, you can prove similarly that $f(x):=(x-a)e^x$ does the job.