Is a projection onto the span of polynomials also an interpolant?

Question

Let $\{p_k\}_{0\leq k\leq N}$ be an orthonormal collection of polynomials. The projection of $f\in L^2([a,b])$ onto $C:=\textrm{span}\{p_k\}_{0\leq k\leq N}$, is denoted (and computed) via

$$p:=\textrm{Proj}_{C}f=\sum_{k=0}^N \langle f \, | \, p_k\rangle p_k.$$

My question:

Let $f\in L^2([a,b])$ be continuous. Does there exists some nonempty collection of nodes $\{x_k\}_{k\in K}\subset[a,b]$ such that, for every $k\in K$, $p(x_k)=f(x_k)$?

Follow-up questions: Are there conditions on $\{p_k\}_{0\leq k\leq N}$ which guarantee this property? Is there a way to determine $\{x_k\}_{k\in K}$ a priori? Is there a characterization on the cardinality of $K$? Is there a classical result, e.g. for Chebyshev polynomials? What if I Gram-Schmidt some Lagrange polynomials?

Thematically, I am asking "When is the projection onto a space of polynomials also an interpolant (in the classical sense of interpolation)?" Any references are greatly appreciated, thank you for your time!

Notes:

1. I'm considering the Hilbert space $L^2$, so I'm really just using $f$ to denote its equivalence class. Also, the values of $f$ are well-defined thanks to continuity.

2. Ideally I would like to find properties of the polynomials and nodes which are independent of $f$.

Answer 1

I believe the answer is "no."

We start with $N$ orthonormal polynomials, $p_k$ ($k = 1, \ldots N$), and continuous $f$. We might as well, by substituting $t = (b-a) s + a$, assuming that the interval $[a, b]$ is actually $[0, 1]$.

Let's let $W = \text{span}(p_1, \ldots, p_N)$, and $P: L^2 \to W$ be the projection operator defined in the question.

The question is then

Does there exists some nonempty collection of nodes $\{x_k\}_{k\in K}\subset[a,b]$ such that, for every $k\in K$, $P(f)(x_k)=f(x_k)$?

We can simplify this to "Is there any point $ 0 \le c \le 1$ with the property that $$ f(c) = P(f)(c) $$ for every continuous $f \in L^2$?" In other words, to show that the answer is "no," it suffices to show that it's "no" for $K = 1$.

Let's pick $c$ to be an arbitrary point of $[0, 1]$. I'll show that there's a continuous function $f$ in $L^2$ such that $f(c) \ne P(f)(c)$. That'll complete the proof.

Here's the idea: no matter what the $p_k$ are, there are tons of functions orthogonal to all of them. At least one of those functions is nonzero at $c$. For that function $P(f)$ is identically zero, so $P(f)(c) = 0$, but $f(c)$ is nonzero.

I'm going to look at something I'll call $C^2$, the space of continuous functions on $[0, 1]$, which is clearly a subset of $L^2$. Define $$ E: C^2 \to \Bbb R : g \mapsto g(c). $$ That's a continuous linear function, and applying it to the function $g(x) = 1$, we see that it's surjective. The kernel of $E$ therefore has codomension $1$.

The orthogonal complement $H$ of $W$ (in $C^2$) has codimension $N$. That means that there's some nonzero vector $g$ in $H$ that's not in the kernel of $E$. [It requires a little linear algebra to see why].

That vector is a continuous function whose projection of $W$ is everywhere zero, but which is not in the kernel of $E$, so whose value at $c$ is nonzero. We're done.

Answer 2

This is an answer to a question in a comment on my earlier answer, but also a self-contained answer to the whole question.

You ask "Any intuition on an explicit counterexample."

Sure, and my solution will follow the ideas in my previous answer. Let's look at $N = 1$, and $p_1(x) = 1$ (i.e., the constant function $1$). $p_1^2$ integrates to $1$ over the unit interval, so $\{p_1\}$ is an orthonormal set of polynomials.

Now I need to look at some functions $f_1$, and the fact that $E$ has kernel dimension $1$ suggests that picking a random one is very likely to get me one that's not in the kernel. Let's give it a shot and pick $f_1(x) = x$.

You can compute that $P(f_1) = \frac{1}{2} p_1(x) = \frac{1}{2}$. There are no surprises here if you draw the pictures.

It's easy to see that $P(f_1)(a) = f(a)$ only for $a = \frac12$. So $a = \frac12$ is the only potential "node" for our set of orthogonal polynomials.

Now let's look at another $f$, say, $f_2(x) = x^2$. For this function, $\langle f_2, p_1 \rangle = \int_0^1 x^2 \cdot 1 \ dx = \frac13$. So $P(f_2) = \frac13$. This is equal to $f_2(x) = x^2$ only if $x^2 = \frac13$, i.e., if $x = \frac1{\sqrt{3}}$. In particular, the functions $f_2$ and $P(f_2)$ do not agree at the only possible "node" that we found for $f_1$. Hence there's no single node for this (very small) set of orthonormal polynomials on the interval $[0, 1]$.