real analysis: uniform convergence of sequence of functions.

Question

Here, we consider everything is defined on $\mathbb R$. Let $f(x) = \lim f_n (x)$, where $f_n(x)$ is sequence of functions. If each of $f_n(x)$ is bounded on $[a, b]$ and $f(x)$ is also bounded on $[a, b]$, then $f_n \to f$ uniformly on $[a, b]$. Here, $a$ and $b$ are real numbers. Is the statement true?

Answer 1

No. Consider $f_n(x)=x^n$ on $[0,1]$. This function converges to $$ f(x)=\begin{cases} 0,& 0\leq x<1 \\ 1,& x=1 \end{cases} $$ which meets all of the criteria. However, the convergence is not uniform. To see why, the definition of uniform convergence is that $\forall \epsilon>0, \exists N$ s.t. $\forall x \in [0,1]$ and $n\geq N,$ $|f_n(x)-f(x)|<\epsilon.$

The negation is $\exists \epsilon_0$ s.t. $\forall n$, $\exists x \in [0,1]$ and $m\geq n$ where $|f_n(x)-f(x)|\geq \epsilon_0.$

Let $\epsilon_0\in (0,1)$ be fixed (any choice works). Then for any $m>n,$ we can choose $x=\epsilon_0^{\frac1m}$. Then $|f_n(x)-f(x)|=|f_n(x)|=\epsilon_0^{\frac{n}{m}}>\epsilon_0.$