I would like to know if the sequence $f_{n}(x)=\frac{1}{e^{xn}}$ converges pointwise and uniformly. $f_{n}:[0, \infty) \rightarrow \mathbb{R}$, for $n\in \mathbb{N}$.
I think it is easier to show that it converges pointwise, because if we take the interval $\tilde{x}=[0,1]$ and assume that if $n \rightarrow \infty$, we would have that $f_{n}(x)=\frac{1}{e^{\tilde{x}n}} \rightarrow 0$. Nevertheless, when it comes to uniform convergence, may I say that $ \sup|f_{n}(x)-0|=\sup_{x\in[0,1]}|\frac{1}{e^{xn}}|=\frac{1}{e^{n}}$?
My point is that this function does not converge uniformly, because in the interval $[0,1]$ we would have $f_{n}(0)=1$ and $f_{n}(1)=\frac{1}{e^n}$.
Is my conclusion correct?
Best Answer
For each $x\in\Bbb R$, you have$$\lim_{n\to\infty}f_n(x)=\begin{cases}1&\text{ if }x=0\\0&\text{ if }x\in(0,\infty).\end{cases}$$Since $(f_n)_{n\in\Bbb N}$ converges pointwise to a discontinuous function and since each $f_n$ is continuous, the convergence is not uniform.