If each $f_n$ is radial Schwartz, and $\|f_n\|_{L^{\infty}} \geq 1,$ can we choose $\{x_n\}$ bounded with $|f_n(x_n)|\geq 1$?

Question

Let $f_n:\mathbb R \to \mathbb C$ be Schwartz class functions, with $\|f_n\|_{L^{\infty}} \geq C$ for all $n$ for some fixed constant $C>0,$ and assume that $f_n(x)= f_n(|x|) $ for all $x$ and $n$.

Question: Can we expect to choose $\{x_n\}\subset \mathbb R$ such that $|x_n|\leq M$ (some $M>0$ fixed )and $|f_n(x_n)|\geq C$ for all $n$?

Answer 1

No. Take a bump function $\varphi$ with support in $(-1,1)$ with $\varphi(0) = C$ and $|\varphi(x)| \leq C$ for $x \neq 0$. Define $$f_n(x) = \varphi(n+x) + \varphi(n-x)$$

Answer 2

The argument proposed by $\color{darkgreen}{user159517}$ is called sliding hump, it is generally used as a counter example of uniform convergence, and since you use $||\cdot||_{L^\infty}$ here, no surprise that it works.

To vizualize it, we can take for instance : $\displaystyle{f_n(x)=e^{-(x-n)^2}+e^{-(x+n)^2}}$

or similarly $\displaystyle{g_n(x)=e^{-(|x|-n)^2}}$

Drawing of the function $f_4$ .

It is the same in $\mathbb R^d$, just replace $|x|$ by $r=\sqrt{\sum\limits_d x_i^2}\quad$ you said yourself it was a radial function.

I searched for a decent 3D-plotter online, found some but wasn't satisfied with the result of the drawing so I figured out that putting a picture of a dog bowl was faster while being almost as accurate.

$\displaystyle{g_n(x,y)=e^{-(r-n)^2}}$