I found some false statements on the uniform convergence and I tried to compute a counter-example for each of them. I would really appreciate is someone could tell me if my counter examples hold, please. The statements are:
Let $f_n$ be a sequence of functioons such that $f_n\to f$ pointwise on a set $E$ as $n\to\infty$
1) If each $f_n$ is continuous on $E$, then $f$ is continuous on $E$ as well.
2) If each $f_n$ is differentiable on $E$, then $f$ is differentiable on $E$ as well
3) Suppose here that $f_n\to f$ uniformly on $E$. If each $f_n$ is differentiable on $E$, then $f$ is differentiable on $E$ as well
Thank you in advance for help!
Counter-examples:
1) Consider $f_n(x)=x^n$ on the interval $[0,1]$. Clearly, $f_n\to f$ pointwise on $[0,1]$ with $f(x)=0$ if $0\le x<1$ and $f(x)=1$ if $x=1$. But, $f$ is not continuous on $[0,1]$. So, the statement doesn't hold.
2) Here we can consider the same $f_n(x)$ as in 1). Each $f_n$ is differentiable on $[0,1]$, but as $f$ is not continuous on [0,1], it is not differentiable on $[0,1]$. (Not really sure about this example)
3) Consider $f_n(x)=\sqrt{x+\frac{1}{n}}$ on $[0,1]$. Clearly, $f_n(x)\to f(x)=\sqrt{x}$ pointwise on $[0,1]$. We can prove now that $f_n\to f$ uniformly on $[0,1]$. $\forall \epsilon>0 \ \forall n\ge N \forall x \in [0,1]$ we have:
$|f_n(x)-f(x)|=|\sqrt{x+\frac{1}{n}}-\sqrt{x}|\underbrace{=\sqrt{x+\frac{1}{n}}-\sqrt{x}}_{\sqrt{x} \ increasing}=\frac{x+\frac{1}{n}-x}{\sqrt{x+\frac{1}{n}}+\sqrt{x}}=\frac{1}{n(\sqrt{x+\frac{1}{n}}+\sqrt{x})}\le \frac{1}{\sqrt{n}}$. Choosing $N=\frac{1}{\epsilon^2}$, we have that $|f_n(x)-f(x)|<\epsilon \ \forall \epsilon>0$. So, $f_n(x)\to f(x)$ uniformly on $[0,1]$. Each $f_n$ is differentiable on $[0,1]$, but $f(x)=\sqrt{x}$ is not differentiable on $[0,1]$ as the derivative on $x=0$ doesn't exist.
Your counterexamples are great. Obviously, as 1) has a counterexample, 2) also as you mentioned.
You can have a look here for more counterexamples on the derivative of sequences of maps.