# Does uniform convergence on the complement of a one point set imply uniform convergence?

Two different function sequences $\{f_n\}$ and $\{g_n\}$ both converge to a function $h : \mathbb{R} \to \mathbb{R}$ uniformly everywhere in $(a,b)$ except at $x = c \in (a,b)$ where they converge non uniformly. I'd like to know if $\{f_n - g_n\}$ converges to zero uniformly everywhere ?

-
This question reads strangely or I am failing to understand something. If the $f_n$ converge uniformly on the two subintervals and pointwise at the point $c$, then they must converge uniformly on the interval $[a,b]$. – ncmathsadist Jun 6 '11 at 2:44
Agreed. The best answer seems to be to point out that there are no sequences of functions $\{f_n\}$ and $\{g_n\}$ satisfying the hypotheses of the question. – Pete L. Clark Jun 6 '11 at 5:12
I think he just means that uniform convergence on $(a,b)\setminus\{c\}$ is hypothesized, and the question is whether this implies uniform convergence on $(a,b)$. – Jonas Meyer Jun 6 '11 at 5:24
@Jonas: well, that's not what the OP said. But if the OP had known that fact he would not have had to ask the question, so I agree that concentrating on that point is most helpful. – Pete L. Clark Jun 6 '11 at 8:32
@Pete : you are right, but now I got to know somehow, then i followed up with another question which is what I actually intended in the first place. – Rajesh Dachiraju Jun 6 '11 at 10:34

I'm not sure how you can have uniform convergence except at a single point. Let $c$ be the point, and let $U=(a,b)\setminus \{c\}$. Because we converge uniformly on $U$ and pointwise at $c$, we have that for all $\epsilon>0$, we have $N_{\epsilon}$ and $M_{\epsilon}$ such that if $n>N_{\epsilon}$, then $|f(y)-f_n(y)|<\epsilon$ for all $y\in U$ and if $n>M_{\epsilon}$, then $|f(c)-f_n(c)|<\epsilon$. However, if we use $N=\max(M_{\epsilon},N_{\epsilon})$, then $|f_n(y)-f(y)|<\epsilon$ for EVERY $y\in (a,b)$ (including $c$) when $n>N$, and therefore we converge uniformly.

Or is that not what you meant by uniform convergence except at $c$?

More generally, we can extend this argument to show that if $U= \displaystyle\bigcup_{1\leq i \leq n} U_i$ and $f$ converges uniformly on each $U_i$, then $f$ converges uniformly on $U$. If $U_i$ is a single point, then pointwise convergence on $U_i$ is the same as uniform convergence.

-
Consider $x^n$ on $[0,1]$. Then consider adding the 'reflection' over $x = 1$ so that the function is defined on $[0,2]$. – mixedmath Jun 6 '11 at 3:06
@mixedmath I agree that in that case the sequence $x^n$ does not converge uniformly, but I don't think you can make it converge uniformly by disregarding a single point in its domain. – gfes Jun 6 '11 at 3:13
@mixedmath That doesn't converge uniformly away from $0$. It converges pointwise to $0$ except at $1$. Let $\epsilon>0$. Note that $e^{-2x}<1-x<e^{-x}$ for $0<x<1/2$. Therefore, for any $n$, we can find an $x$ close to $1$ such that $x^n>\epsilon$ (and the formula allows us to find such an $x$ explicitly). – Aaron Jun 6 '11 at 3:26
@mixedmath And since I already gave a proof that the behavior cannot happen, you are going to be very hard pressed to find an example. – Aaron Jun 6 '11 at 3:30
You know, for some reason I had in my mind the idea that this was the classic example of some weird behavior, so I didn't even check. But you are absolutely correct. +1. – mixedmath Jun 6 '11 at 13:02

Actually if $f_n \rightarrow h$ on $(a,b)$ and uniformly on $(a,c)\cup(c,b)$, then $f_n \rightarrow h$ uniformly on $(a,b)$.

To see this, we invoke the definition of uniform convergence $f_n \rightarrow h$: $\forall \epsilon>0: \exists N_\epsilon: n\geq N_\epsilon \Rightarrow |f_n-h|_\infty<\epsilon$

If this holds on $(a,c)\cup(c,b)$, and convergence holds on $c$: $\forall \epsilon>0: \exists M_\epsilon: m\geq M_\epsilon \Rightarrow |f_m(c)-h(c)|<\epsilon$

Then picking $K_\epsilon = \max(N_\epsilon,M_\epsilon)$ establishes: $\forall \epsilon>0: \exists K_\epsilon: k\geq K_\epsilon \Rightarrow |f_k-h|_\infty<\epsilon$

on $(a,b)$ and therefore proves uniform convergence on $(a,b)$.

So we have $f_n$ and $g_n$ both converging uniformly to $h$ on $(a,b)$. Thus: $\forall \epsilon>0: \exists K_\epsilon: k\geq K_\epsilon \Rightarrow |f_k-h|_\infty<\epsilon$ and $\forall \epsilon>0: \exists L_\epsilon: l\geq L_\epsilon \Rightarrow |g_l-h|_\infty<\epsilon$

Now picking $P_\epsilon = \max(K_\frac{\epsilon}{2},L_\frac{\epsilon}{2})$ gives:

$\forall \epsilon>0: \exists P_\epsilon: p\geq P_\epsilon \Rightarrow |f_p-h|_\infty<\frac{\epsilon}{2}$ and $|g_p-h|_\infty<\frac{\epsilon}{2}$ and thus by the triangle inequality $|(f_p-h)+(h-g_p)|_\infty = |(f_p-g_p)-0|_\infty < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$

So we've proved uniform convergence of $f_p-g_p$ to $0$ on $(a,b)$.

-