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Consider the sequence of continuous functions $f_n(x) : mathbb{R} to mathbb{R}$
defined by $f_n(x) = frac{x}{n}$. In which of the following topologies: pointwise convergence, compact convergence, and uniform does $f_n(x)=frac{x}{n}$ converge. What about when $f_n(x) = frac{1}{n^3[x-(1/n)]^2 + 1}$?

We define the uniform topology on $Y^X$={functions from X to Y} as the topology with subbasis ${B(f,ϵ):fin Y^X,epsilon>0}$, where
$B(f,ϵ)={g∈Y^X:text{sup }_{x∈X} d(f(x),g(x))<ϵ}$

and the compact convergence topology is generated by the collection of sets
$B_C(f,ϵ)={g∈Y^X:text{sup}_{x∈C} d(f(x),g(x))<ϵ}$,
with compact $C⊂X$

and the pointwise convergence topology is generated by the subbasis $S(X,U) = {f|f in Y^X text{ and } f(x) in U }$ for $U$ open in $Y$.

Let us recall the definitions of pointwise, uniform and uniform on compact convergence.

Given functions $f_n : mathbb R to mathbb R$ and $f : mathbb R to mathbb R$, we say that :

• $f_n to f$ pointwise if for each $x in mathbb R$ we have $f_n(x) to f(x)$.




• $f_n to f$ uniformly, if for all $epsilon > 0$ there exists $N in mathbb N$ such that if $n > N$ then for all $x$ we have $|f_n(x) - f(x)| < epsilon$.




• $f_n to f$ uniformly on compact sets, if for each compact set $K subset R$, the restricted functions $f_n|_K$ and $f|_K$ satisfy $f_n|_K to f|_K$ uniformly on $K$.

With this in mind, for $f_n = frac xn$, we need a candidate function for $f$, where $f_n$ would converge pointwise. But, for each fixed $x$, the sequence $frac xn$ clearly goes to zero. So, $f equiv 0$ should work.

To see this, fix $x in mathbb R$. The point is, for pointwise convergence we are supposed to show for each fixed $x$ that $f_n(x) to f(x)$, while in the uniform case we are supposed to specify $N$ before fixing $x$. That is the difference in the two forms of convergence.

But if $x$ is fixed, then the sequence $frac xn$ clearly converges to $0$. Since this is true for all $x$, we see that $f_n$ converges pointwise to $0$.

But not uniformly Why? The intuition behind uniform convergence, is that in some sense, the rate at which $f_n(x) to f(x)$ happens, should be the same for each $x$ (or at least uniform in some sense). Here, this is not the case, and we argue by contradiction. Suppose that $f_n to f$ uniformly. Then, $f_n to f$ pointwise(WHY?), so $f equiv 0$ must happen. We know that for all $epsilon > 0$ there exists $N$ such that $n > N$ implies $|f_n(x)| < epsilon$ for all $x$. But this is false : let $epsilon = 1$ . Then, if $N$ comes out from the definition of uniform continuity, take $x = N+2$ to see that $f_{N+1}(N+2) > 1 = epsilon$, so it is not true that if $n > N$ then $|f_n(x)| < epsilon$ for all $x$, as we showed by taking a specific $n$ and $x$.

In short, the rate of convergence to zero of $f_n(x)$ is not uniform : in fact, it gets slower as $x$ grows larger.

Uniformly on compacts? Let $K$ be compact. The candidate for uniform convergence on $K$ should again be the zero function. Let us fix $epsilon > 0$. We are supposed to find $N$ such that if $n > N$ then $|f_n(x)| < epsilon$ for all $x in K$. Expanding, this is saying that $|x| < nepsilon $ for all $x in K$.

But a compact set is bounded! So let $M$ be such that $|x| < M$ for all $x in K$. Simply pick $N > frac{M}{epsilon}$ an integer. Then, if $n > N$ we have $nepsilon > M > |x|$, or that $|f_n(x)| < epsilon$. Hence, uniform convergence on compacts is true here!

Use the ideas above to solve the second question.

• First, find a candidate for pointwise convergence. If the convergence is uniform/uniform on compacts, then the candidate remains the same.




• Study the values of $f_n(x)$ for various values of $x$ to check intuitively whether uniform convergence applies. Study $|f_n(x) - f(x)|$ for varying $n$ and $x$ : this gives you a good clue.




• For compactness, you have to see if boundedness makes a difference to the rate of convergence.Usually, if $|f_n(x) - f(x)|$ has a bound related to an upper bound of the compact set, then this is helpful.