relaxation on Dini's Theorem

Question

Dini's Theorem states: Let $[a,b]$ be a compact intervall. Let $f,f_{n}: [a,b] \xrightarrow{} \mathbb{R}$, $ n \in \mathbb{N}$, functions with

1. $f$ and $f_{n}$ are continuos for all $n$
2. $\lim_\limits{n \to \infty} f_{n}(x) = f(x)$ $\forall x \in [a,b]$
3. $f_{n}(x) \leq f_{n+1}(x)$ $\forall x \in [a,b]$ and $n \in \mathbb{N}$.

Then $\{f_{n}\}_{n \in \mathbb{N}}$ converges uniformly to $f$.

Suppose only conditions $1$ and $2$ are satisfied. Are in this case $\{f_{n}\}_{n \in \mathbb{N}}$ also converges uniformly to $f$?

The reason why I'm asking this quesion ist that I found a variant of Dini's theorem that states if condition $1$ and $2$ are satisfied and we have that $f_{n}$ is a so called commute sequence, i.e. f_{n} has a monotonically increasing and a monotonically decreasing part, then Dini's Theorem also holds

Answer 1

Let $f_n(x)=nx$ for $x \in [0,\frac 1 n]$ and $f_n(x)=2-nx$ for $x \in [\frac 1 n, \frac 2 n]$, $0$ in the rest of $[0,1]$. Then $f_n \to 0$ pointwise but not uniformly since $f_n(\frac 1 n) =1$ for all $n$.

Answer 2

Consider $f_n : [0,1] \to \mathbb{R}$ defined as $f_n(x) = nxe^{-nx}$. Clearly $f_n \to 0$ pointwise, but $$f\left(\frac1n\right) = e^{-1}, \forall n\in\mathbb{N}$$

so the convergence is not uniform.