Understanding uniform convergence of sequence

Question

Given $f_n(x)=(x-1)^{3n}$ on $(0,2]$ we can show point-wise convergence to the function $$ f(x)=\begin{cases} 0&x\in(0,2)\\1&x\in\{2\}\end{cases} $$ But how do I show that uniform convergence is not true? I've tried the following:

For $x\in(0,2)$ we can show point wise convergence of $f_n(x)$ to $f(x)=0$ by observing that $$ |(x-1)^{3n}|\le|x-1|^{3N} $$ for all $n\ge N$. Let $N=\lceil\frac{1}{3}\frac{\log\epsilon}{\log|x-1|}\rceil$. For all $\epsilon>0$ we have that $$ |(x-1)^{3n}|\le|x-1|^{3N}=\epsilon $$ for all $n\ge N$. Since $N(x,\epsilon)$ is a function of $x$, $N$ is not chosen independently of $x$ and thus uniform convergence is not true on $(0,2]$.

Answer 1

Uniform convergence of a sequence $f_n(x)$ defined in some interval $I$ to a function $f(x)$ is equivalent to

$$\lim_{n\to\infty}\sup_{x\in I}|f_n(x)-f(x)|=0\quad\quad (1)$$

Indeed, if $f_n\to f$ uniformly on $I$, then for every $\varepsilon>0$ we can find $N$ such that for every $n>N$ and every $x\in I$, $|f_n(x)-f(x)|<\varepsilon$. Since this holds for every $x\in I$, it holds for the supremum too, hence $\sup_{x\in I}|f_n(x)-f(x)|<\varepsilon$ for every $n>N$, hence $(1)$ holds. It is equally easy to see that if $(1)$ holds, then $f_n\to f$ uniformly on $I$.

This implies in particular that if you can find a sequence $x_n\in I$ such that $f_n(x_n)-f(x_n)$ is not close to zero, then the convergence cannot be uniform. In our case, if you take $x_n=2-\frac{1}{n}$, you find: $$f_n(x_n)-f(x_n)=(2-\frac{1}{n}-1)^{3n}\to \frac{1}{e^3}\quad\hbox{as $n\to\infty$}$$ Consequently, $(1)$ does not hold, hence $f_n$ does not converge to $f$ uniformly.

Answer 2

Simple. There's a standard theorem that a uniform limit of continuous functions is continuous. These functions are continuous, but their limit is not. Therefore it is not a uniform limit.

The proof of that theorem is quite short, so if you need to translate it to your specific function, that should be easy. It's called the "epsilon-over-three argument", and is one of the classic proofs. See for example https://en.wikipedia.org/wiki/Uniform_limit_theorem

Answer 3

One simple way of proving it is to use this theorem that says:

If a sequence of continuous functions $f_n$ converges to a function $f$ uniformly, then $f$ is continuous.

Using this theorem, it's clear that since $f_n$ is continuous, and $f$ is not, convergence cannot be uniform.

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However, in your case, you haven't proven that the functions don't converge uniformly. You have only proven that one particular way of choosing $N$ is not independent of $x$. What you need to do is prove that every way of choosing $N$ must depend on $x$.

If you want to show that convergence isn't uniform directly, there is no way around the first step which is checking the definition of uniform continuity. The definition says:

A sequence of functions $f_n:(a, b]$ converges to $f$ uniformly if, for every $\epsilon>0$, there exists some $N$ such that, for all $x\in (a,b]$, the inequality $|f_n(x)-f(x)| < \epsilon$ is true.

Using this definition, we can of course write its negation to get

A sequence of functions $f_n$ does not converge to $f$ uniformly if there exits some $\epsilon > 0$ such that, for all $N\in\mathbb N$ , there exits some $x$ such that $|f_n(x)-f(x)|\geq\epsilon$.

In your case, I advise you to look at the numbers $x_k = 2-\frac1k$ and look at $|f_n(x_k) - f(x_k)|$. No matter how big $n$ is, you can choose a large enough $k$ (and, therefore, an appropriate $x_k$) for that absolute value to be quite big.

Answer 4

The fact that your $N$ is not chosen independently of $x$ does not prove yet that there is no better choice. So actually you did not prove it in your attempt.

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Suppose that there is uniform convergence.

Then there must be a positive integer $n_0$ such that $n>n_0$ implies that $|f_n(x)-f(x)|<0.3$.

Then consequently $|f_n(x)|\leq0.3$ for $x\in(0,2)$.

But next to that we have $f_n(2)=1$ so this contradicts the continuity of $f_n$ (which requires that $\lim_{x\to 2^-}f_n(x)=f(2)$).

This contradiction allows us to conclude that there is no uniform convergence.

Answer 5

Your argument sounds right to me. In fact the uniform convergence screws up when $x=2$ or $x\to 0$. Now by contradiction assume uniform convergence holds. Then we must have $$\forall\epsilon>0\qquad\exists M\qquad \forall n>M,0<x\le 2\qquad |f_n(x)-f(x)|<\epsilon$$this means that $$\forall n>M\qquad\text{if }x = 2-\zeta\text{ for some }0<\zeta<1\to|f(x)|<\epsilon$$therefore $$(1-\zeta)^{3n}<\epsilon$$or $$n>\dfrac{\log_{1-\zeta}\epsilon}{3}=\dfrac{\log_{x-1}\epsilon}{3}=M$$therefore $M$ must depend on $x$ and this is a contradiction.