Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
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
up vote 5 down vote accepted

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.

share|cite|improve this answer
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

The short answer is yes.

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 $(a,b)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.