There exists $\{f_n\}$ as above that converges to $0$ uniformly on $[0,1]$.

Question

Q. Let $\{f_n\}$ be a sequence of functions defined on $[0,1]$. Suppose there exists a sequence of distinct numbers $x_n\in[0,1]$ such that $$f_n(x_n)=1.$$ Prove or disprove the following statements:

a) True or false: There exists $\{f_n\}$ as above that converges to $0$ pointwise.

False: Pick $\varepsilon=1/2$, and $x=x_n$, and $M\in N$. Then, for $n\ge M,$ $$|f_n(x_n)-0|=1<\varepsilon=1/2.$$This is a contradiction.

b) True or false: There exists $\{f_n\}$ as above that converges to $0$ uniformly on $[0,1]$.

False: $\lim_{n\to \infty}||f_n-0||=\lim_{n\to \infty}\sup(f_n-0)\ge\lim_{n\to \infty}1\ne0.$

I am not sure about a). Could you check this?

Answer 1

You are wrong about a). Take$$f_n(x)=\begin{cases}2nx&\text{ if }x\in\left[0,\frac1{2n}\right]\\2-2nx&\text{ if }x\in\left[\frac1{2n},\frac1n\right]\\0&\text{ otherwise.}\end{cases}$$Then $(f_n)_{n\in\mathbb N}$ converges pointwise to the null function, but $(\forall n\in\mathbb{N}):f_n\left(\frac1{2n}\right)=1$.