# Confusion over L2-norm polynomial approximation

Q1: Let $$C(\mathcal{X})$$ be the space of continuous functions $$f:[a,b]\to\mathbb{R}$$. Let $$F(\mathcal{X})$$ be a subset of $$C(\mathcal{X})$$ that is dense in $$C(\mathcal{X})$$ w.r.t. metric $$d_{\infty}(\cdot,\cdot)$$. Is $$B=\{f\in F(\mathcal{X}): \lVert{f}\rVert^2\leq 1\}$$ dense in $$A=\{f\in C(\mathcal{X}): \lVert{f}\rVert^2\leq 1\}$$? ($$\lVert{\cdot}\rVert^2$$ is L2-norm.)

Q2: Assume the answer for Q1 is yes. Let $$B,A$$ be defined as Q1. For two points $$x_1,x_2\in [a,b]$$, we have \begin{align*} \sup_{f\in A} \lvert f(x_1)-f(x_2)\rvert =\sup_{f\in B} \lvert f(x_1)-f(x_2) \rvert.\tag{*} \end{align*}

Let $$F(\mathcal{X})=\overline{\mathrm{span}}\{k_x: x\in[a,b]\}$$, where $$k_x$$ is a feature map induced by a Gaussian kernel $$\langle k_x,k_y\rangle=k(x,y)=\mathrm{e}^{{-\alpha(x-y)}^2}$$. Then $$F(\mathcal{X})$$ is a RKHS and it has been proven that it is dense in $$C(\mathcal{X}$$) for any positive constant $$\alpha$$. Then I get: \begin{align*} \sup_{f\in B} \lvert f(x_1)-f(x_2) \rvert &=\sup_{f\in B} \langle k_{x_1}-k_{x_2}, f\rangle\\ &=\sqrt{\langle k_{x_1}-k_{x_2},\ k_{x_1}-k_{x_2}\rangle}\quad\text{ (**)}\\ &=\sqrt{2-2\mathrm{e}^{{-\alpha(x_1-x_2)}^2}}. \end{align*} Now the result depends on $$\alpha$$. It is expected to get a constant value for different $$\alpha$$. What's wrong with the process?

• What norm do you use in the definition of $A,B$? If you use the sup-norm, then the answer is yes to Q1.
– daw
May 3, 2021 at 9:58
• It's actually L2 norm. Is the claim still true? May 3, 2021 at 10:43

## 1 Answer

I think you mix up the two different(!) norms $$\lVert f\rVert_{L^2}$$ and the norm from the Gaussian Kernel $$\lVert f\rVert_{\exp}$$. Your equation (*) assumes you work with the unit balls of the $$L^2$$-norm while (**) is only true for the unit ball in the RKHS norm.

With respect to (**): Let $$\langle\cdot,\cdot\rangle_{\exp}$$ be the scalar product of the Gaussian RKHS. Then the equation should read/is only true for: $$\sup_{\lVert f\rVert_{\exp}\leq 1} \langle k_{x_1}-k_{x_2}, f\rangle_{\exp} =\sqrt{\langle k_{x_1}-k_{x_2},\ k_{x_1}-k_{x_2}\rangle_{\exp}}.$$

And since $$\lVert f\rVert_{L^2} \neq \lVert f\rVert_{\exp}$$ the unit balls $$B= \{f\mid \lVert f\rVert_{L^2} \leq 1\}$$ and $$\{f\mid \lVert f\rVert_{\exp} \leq 1\}$$ are not equal. That is why $$\sup_{f\in B} \langle k_{x_1}-k_{x_2}, f\rangle_{\exp} \neq \sqrt{\langle k_{x_1}-k_{x_2},\ k_{x_1}-k_{x_2}\rangle_{\exp}}.$$

(And $$\alpha$$ has to be negative)

• Thank you for the answer. I'm still a bit confused. $F(\mathcal{X})$ induced by Gaussian kernel is dense in $C(\mathcal{X})$ in terms of $d_\infty(\cdot,\cdot)$. Both the elements in $A$ and $B$ are restricted to be from unit ball (L2-norm). The functions in $A$ uniformly converge to a function in $B$. That's why the equation $(∗)$ holds. $(∗∗)$ is just a solution for the RHS of $(∗)$. The problem is that the solution changes with $\alpha$, which is supposed to be a constant according to equation $(∗)$. May 3, 2021 at 11:08
• see my edits I hope it is clearer now.
– g g
May 3, 2021 at 12:09