uniformly and pointwise

Question

Consider the sequence of functions $(f_n)$ defined on $[0,1]$ by $f_n(x) = x^n$. Show that the sequence converges to the function

$f(x) = 0 $ if $ x ∈ [0,1)$ and $f(x)=1$ if $x = 1$

pointwise on $[0,1]$, but not uniformly on $[0,1]$.

Answer 1

Pretty much the standard deal, if $|x|<1$, then $$ \lim_{n\rightarrow \infty}x^n=0 $$ since it is a monotone decreasing sequence bounded below.

If $x=1$, $\lim_{n\rightarrow \infty}x^n=1$.

As the limit is not a continuous function, and each $f_n$ is continuous, the convergence cannot be uniform.

Answer 2

It is not difficult to note that the sequence converges pointwise to $f$. Surely, set $x=1$ and you get a constant sequence. And, for any $x\in[0,1)$, the sequence converges monotonically to $0$. I am not including the details of the proof.

Now, observe that , for any $x_0\in(0,1)$ $$|x_0|^n<\epsilon \text{ iff. } n>\dfrac{\ln\epsilon}{\ln x_0}$$

Assume that the sequence was uniformly convergent. So, for any $x_0\in(0,1)$ $$n>K\implies|x_0|^n<\epsilon$$ for some $K.$ And, since the choice of $x_0$ is arbitrary, it is necessary that $$K\ge \dfrac{\ln\epsilon}{\ln x_0}$$

But, it should be noted that $$\lim_{x_0\to1^-}\dfrac{\ln \epsilon}{\ln x_0}=\infty$$ which poses a contradiction. So, it can't be uniformly convergent.