Uniform convergence with two limits

Question

I'm doing a question investigating uniform convergence of a function and I need something cleared up if possible.

$f_n(x) = \frac{x^n}{1+x^n}$ on the interval $[0,1]$.

Now, pointwise, this turns into a piecewise limit function:

$$f(x) = \left\{\begin{array}{ll}{1\over2}&\text{ if }x=1 \\ 0&\text{ if }x\neq 1\end{array}\right.$$

Anyways, provided this limit function is correct, how does one prove or disprove uniform convergence here, using the $\displaystyle \lim_{n\rightarrow \infty} \sup_{x\in I} |f_n(x)-f(x)|=0$ definition?

Thanks!

Answer 1

Fill in the expression $$\lim_{n\to\infty}\sup_{x\in I}|f_n(x)-f(x)|$$ from the inside out:

(1) Plug in the limit function $f(x)$ into $|f_n(x)-f(x)|$. The result will be $|f_n(x)|$ if $x\ne1$, and will be zero if $x=1$. So the expression is now

$$\lim_{n\to\infty}\sup_{x\ne1}|f_n(x)|$$

(2) Calculate, for each $n$, the sup of $|f_n(x)|$ over $x\ne1$. Check that the answer is $1/2$ for each $n$.

(3) Now the expression is $$\lim_{n\to\infty}1/2\;,$$ which equals $1/2$.

Answer 2

For any $n$ we have $\sup_{x\in I} |f_n(x) - f(x)|=1/2$ (to see this, look at $|f_n(x) - f(x)|$ as $x\to 1$). Thus $\lim_{n\to\infty}\sup_{x\in I} |f_n(x) - f(x)|=1/2$.

Answer 3

The contradiction comes from the fact that $(C[0,1], ||\cdot ||_{\infty})$ is closed, i.e. If a sequence of continuous functions converges uniformly, then its limit is continuous.

Observe that $f_n(0)=0$, $f_n(1)=\frac{1}{2}$. Since $f_n$ is continuous, then for every $n \in \mathbb N$ there exists a certain $\xi_n<1$ such that $f(\xi_n)=\frac{1}{4}$. This implies that

$$\sup_{x\in [0,1]}|f(x)-f_n(x)|\geq |f(\xi_n)-f_n(\xi_n)|=f_n(\xi_n)=\frac{1}{4}$$

This implies that the limit is greater than $0$.