Test the sequence of functions $$x_n(t)={e^{-nt}}$$ for convergence in $C[0,1]$,$L_{1}(0,1)$ and $L_{2}(0,1)$. In case of convergence, find the limit function.
What I have done.
(1) In $C[0,1]$
For $C[0,1]$ we have that $\lim_{n->\infty} e^{-nt}$ converges to $0$ when $0<t\leq1$, and to $1$ when $t=0$.
Can I say, without further testing, that it diverges in $C[0,1]$, because it does not converges uniformelly?
(2) In $L_{1}(0,1)$
Point-wise convergence: In this case, because $t \neq 0$, the sequence $$\lim_{n->\infty} e^{-nt}=0$$ when $t \in (0,1)$ is fixed. It converges point-wise on $(0,1)$
Uniform convergence: I computed the norm of this metric space $$ \|x\|_{L^1} = \int_{0}^1|x(t)|\,dt =\frac{1-e^{-n}}{n} $$.
Now, the sequence $\frac{1-e^{-n}}{n} $ converges to $0$ when $n$ aproaches $\infty$, then $x_n(t)$ converges in $L_{1}(0,1)$. How do I find the limit of the sequence?
(1) Yes, your argument is justified.
(2) The limit is the zero function. $L^{1}$ consists of equivalence classes of functions and values at a finite number of points do not matter.
(3) is similar to (2).