We have points $x_0=a \lt x_1 \lt x_2 ....x_n=b $ and $\;w_{n+1}(x)=\prod_{k=0}^{n}{(x-x_k)}$.
Let $h=max_{j=0...n}|x_j-x_{j-1}|$

Let $f \in C^{n+1}[a;b]$ and $p_n\in \mathbb P_n$ be the corresponding (unique) interpolation polynomial
Prove that $ ||f-p_n||_{\infty}\lt \frac{h^{n+1}}{4(n+1)}||f^{n+1}||_{\infty}$ holds for every $a\le x \le b$

We can assume that $x \in [x_i;x_{i+1}]$ for some $i \in \mathbb N$. Then I would like to get a proper upper bound for $|x-x_k|.$ which I am struggling to find.

My guess would be that $|x-x_k|\lt |x_i-x_k| \lt |k-i|*h)$ but I don't know how that would help me.

Would appreciate some help.


  • $\begingroup$ this is the error function of Lagrange interpolation? $\endgroup$ – cactus314 Jun 11 '16 at 16:03
  • $\begingroup$ It is. I am going to edit my post $\endgroup$ – XPenguen Jun 11 '16 at 16:05

Let $x \in [x_0,x_1]$. Then $|x-x_0| \leq h$ and $|x-x_1| \leq h$. You also have $|x-x_j| \leq jh$ for $j=2,3,\dotsc,n$. It follows, that $$|w_{n+1}(x)| \leq (h)(h)(2h)(3h)\dotsc(nh) = (n!) h^{n+1}.$$ This is an estimate which is compatible with the conclusion which you want to achieve. It remains to generalize this to arbitrary $x$.

I will do the case of $x \in [x_1,x_2]$ and leave the rest to you.

It is clear that $|x-x_1| \leq h$ and $|x-x_2| \leq h$. Ignoring $x_0$ for now we instead move towards the distant point of $x_n = b$! We have $$|x-x_j| \leq (j-1) h, \quad j=3,4\dots,n.$$ It is true that we have $|x-x_0| \leq 2h$, but I will generously overestimate and write $|x-x_0| \leq nh$. Then, as before, we have $$|w_{n+1}(x)| \leq (nh)(h)(h)(2h)(3h)\dotsc((n-1)h) = (n!) h^{n+1}.$$

In general, you move towards the right hand endpoint $b$ and then wrap around.


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