Bound for the max value of a Lagrange polynomial

Question

Given some Lagrange polynomial $L$(x) that interpolates over the points $x$₀, $x$₁,..., $x$_n with values in the set $A$ = {$a$₀, $a$₁,..., $a$_n } on some interval [a,b], show that the max value that $L$(x) takes on the interval is bounded as such:

||$P$||_∞ ≤ M||$A$||₂
where M is a function of "n" (the number of interpolation points).

My Ideas:

||$P$||_∞ is just the magnitude of some point $P$(x_i) (maybe not unique) for $P$(x) on [a,b]. So I started by representing this maximum as the linear combination of the base Lagrange polynomials:

$P$(x_i) = a_oq_o(x_i) + ... + a_nq_n(x_i)

I then squared both sides:

($P$(x_i))² = (a_oq_o(x_i) + ... + a_nq_n(x_i))²

Next, I used the Cauchy inequality to say:

($P$(x_i))² = (a_oq_o(x_i) + ... + a_nq_n(x_i))² ≤ (a_o² +...+ a_n²)(q_o(x_i)² +...+ q_n(x_i)²).

This means:

$P$(x_i)) ≤ ||$A$||₂ (q_o(x_i)² +...+ q_n(x_i)²)^1/2

But I don't know how to turn (q_o(x_i)² +...+ q_n(x_i)²)^1/2 = M into something that's only based on "n".

Answer 1

You are very close. $q_i(x)^2$ is a polinomial that equals $1$ at $x=x_i$ and has a double root at $x=x_j$ for any $j\neq i$, hence $$ q_i(x)^2 \leq \frac{K}{K+(x-x_i)^2} \tag{1}$$ where $K$ just depends on the mesh $\delta$ of the partition given by $x_0,x_1,\ldots,x_n$.

A continuous analogue is given by the inequality: $$\left(\frac{\sin (\pi x)}{\pi x}\right)^2 \leq \frac{\frac{3}{\pi^2}}{\frac{3}{\pi^2}+x^2} $$ where the LHS is a function that equals $1$ at $x=0$ and has a double root at any other integer.

Now, since the function $\frac{K}{K+x^2}$ is integrable over $\mathbb{R}$ and its integral equals $\pi \sqrt{K}$, it is not difficult to provide a uniform bound for $\sum_{j=0}^{n}q_j(x)^2$ that depends on $n$ only.

Anyway, I am waiting to know if some extra information on the partition given by $x_0,x_1,\ldots x_n$ is available. I can give more explicit bounds for uniform or Chebyshev partitions. I am expecting that something has to be known: for instance, by taking $n=2$, $[a,b]=[0,1]$ and $x_i=\frac{5+i}{12}$ we have the following wild behaviour of $q_0(x)^2+q_1(x)^2+q_2(x)^2$:

$\hspace{1in}$

At least we have to know that $x_0$ is close to $a$ and $x_n$ is close to $b$: otherwise, $(1)$ may fail, as depicted above.