Showing that a sequence does not converge uniformly to a piecewise function

Question

I understood how to show that a sequence converge pointwise and uniformly to another function. However, I am finding it difficult to show that a function DOES NOT converge uniformly to a function, in particular when the pointwise limit is piecewise. Let me give an example:

Let $f_n(x)=\dfrac{x}{1+x^n}$ for $n\in \mathbb{N}$ and $x\in[0, \infty)$

I am required to find the pointwise limit, which I found to be:

$$ f(x)= \begin{cases} x \ , \ x\in[0, 1)\\ \frac{1}{2} \ , \ x=1\\ 0 \ , \ x\in(1, \infty) \end{cases} $$

Now for the second part of the question, I am required to show that it does not converge uniformly. I understand that the reason for this (intuitively) is because there is a sort of "jump" at $x=1$

My attempt was as follows:

Let $x=1+\frac{1}{n}$

Now I try to show that for any $N\in\mathbb{N}$ , $\exists n\geq N$, such that $|f_n-f|>\epsilon$ for some $\epsilon$.

Let $\epsilon=1/4$ and let $n>\dfrac{1}{\sqrt[N]{3}-1}$

$$|f_N(x)-f(x)|$$ $$=\left|\dfrac{x}{1+x^N}-0\right|$$ $$=\left|\dfrac{x}{1+x^N}\right|$$ $$=\left|\dfrac{1+\frac{1}{n}}{1+(1+\frac{1}{n})^N}\right|$$ $$>\dfrac{1}{1+(1+\frac{1}{n})^N}$$ $$>1/4=\epsilon$$

First of all I am not sure if this method is correct, and even if it is, is this the best way to go about showing uniform convergence, at least in this case?

I am aware that uniform convergence occurs iff $\lim_{n\rightarrow \infty}\sup{|f_n(x)-f(x)|}=0$, but I cannot see how to use this since the pointwise limit is a piecewise function.

Answer 1

Each $f_n$ is continuous, but $f$ isn't. Therefore, the convergence is not uniform.

Answer 2

For a direct proof that the convergence can't be uniform, note that if $n$ is large enough that $|f_n(1) - 1/2| < 1/4$, and thus $f_n(1) > 1/4$, then by continuity of $f_n$ there is $\delta > 0$ so that $f_n(x) > 1/4$ for $1 < x < 1+\delta$. But $f(x) = 0$ so $|f_n(x)-f(x)| > 1/4$ for such $x$.