If, for every $n$, $f_n\to0$ at infinity and $|f_n(x_n)|\geq C$, can one deduce that $(x_n)$ is bounded? Let $f_n:\mathbb R \to \mathbb C$ such that $f$ is continuous on $\mathbb R$ and $\|f_n\|_{L^{\infty}(\mathbb R)} \geq C &gt;0$ for all $n\in \mathbb N$ (where $C&gt;0$ is fixed constant)

Therefore, we have

(1) There exists $\{x_n\}\subset \mathbb R$ such that  $$|f_n(x_n)| \geq C $$ for all $n\in \mathbb N.$

Assume that $|f_n(x)| \to 0$ as $|x| \to \infty.$


  Question: Can we say $\{x_n\}$  (as obtained in (1) above) is bounded? In other words, Dose there exists $M&gt;0$ such that $|x_n|\leq M$ for all $n$, and $|f_n(x_n)| \geq C$?
This is false in general. 

This can happen: Take something like a piecewise affine function on $\mathbb R$ which is $0$ outside $(0,1/n)$, such that $f(1/2n) = C$ and extends linearly on $(0,1/n)$. This is continuous and there is pointwise convergence to $0$ everywhere. The sequence $1/2n$ is bounded.

Now, notice that the same construction with the interval being $(n,n+1)$ gives you a counter-example.
This is false in general. 

This can happen: Take something like a piecewise affine function on $\mathbb R$ which is $0$ outside $(0,1/n)$, such that $f(1/2n) = C$ and extends linearly on $(0,1/n)$. This is continuous and there is pointwise convergence to $0$ everywhere. The sequence $1/2n$ is bounded.

Now, notice that the same construction with the interval being $(n,n+1)$ gives you a counter-example.