Vtyjkyuk

Given $f_n(x)=(x-1)^3n$ on $(0,2]$ we can show point-wise convergence to the function
$$
f(x)=begincases 0&xin(0,2)\1&xin2endcases
$$
But how do I show that uniform convergence is not true? I've tried the following:

For $xin(0,2)$ 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 $nge N$. Let $N=lceilfrac13fraclogepsilonx-1rceil$. For all $epsilon>0$ we have that
$$
|(x-1)^3n|le|x-1|^3N=epsilon
$$
for all $nge N$. Since $N(x,epsilon)$ is a function of $x$, $N$ is not chosen independently of $x$ and thus uniform convergence is not true on $(0,2]$.

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

$$lim_ntoinftysup_xin I|f_n(x)-f(x)|=0quadquad (1)$$

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

This implies in particular that if you can find a sequence $x_nin I$ 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 $x_n=2-frac1n$, you find:
$$f_n(x_n)-f(x_n)=(2-frac1n-1)^3nto frac1e^3quadhboxas $ntoinfty$$$
Consequently, $(1)$ does not hold, hence $f_n$ does not converge to $f$ uniformly.

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.

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 $xin (a,b]$, 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 $Ninmathbb N$ , there exits some $x$ such that $|f_n(x)-f(x)|geqepsilon$.

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.

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.

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 $xin(0,2)$.

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

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

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

Your argument sounds right to me. In fact the uniform convergence screws up when $x=2$ or $xto 0$. Now by contradiction assume uniform convergence holds. Then we must have $$forallepsilon>0qquadexists Mqquad forall n>M,0<xle 2qquad |f_n(x)-f(x)|<epsilon$$this means that $$forall n>Mqquadtextif x = 2-zetatext for some 0<zeta<1to|f(x)|<epsilon$$therefore $$(1-zeta)^3n<epsilon$$or $$n>dfraclog_1-zetaepsilon3=dfraclog_x-1epsilon3=M$$therefore $M$ must depend on $x$ and this is a contradiction.