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Let $\{f_v\}_v\in\mathbb N$ be a sequence of continuous functions $f_v:\Re^m\to\Re^n$ and $f:\Re^m\to\Re^n$ and assume that $f_v$ converges to $f$ pointwise (i.e. For every fixed $x\in\Re^m$, $f_v(x)\to f(x)$). If the sequence $f_v$ was equicontinuous, then it would also converge uniformly to $f$. I am looking for some alternative conditions that may render $f_v$ also uniformly convergent to $f$.

Consider the continuous functions $f_v(x)=x^v$, where $x\in[0,1]$. These functions converge pointwise to $f(x)=1-\chi_{[0,1)}$ but they do not converge uniformly (and of course they are not equicontinuous). I notice additionally that their pointwise limit is not continuous.

Consider now the following alternative conditions:

  1. $f_v:\Re^m\to\Re^n$ converges point-wise to $f$
  2. $f_v$ are continuous
  3. The sequence $f_v$ is uniformly bounded, i.e. there is a compact set $K\subset \Re^n$ such that $f_v(x)\in K$ for all $v\in \mathbb N$ and for all $x\in\Re^m$.
  4. $f$ is continuous!

Under these assumption, can we show the following property for the sequence $f_v$ (referred to as continuous convergence):

Fix a $x_0\in\Re^m$. For every sequence $x_v\to x_0$, it holds that $f_v(x_v)\to f(x)$.

Can we also show that the sequence $f_v$ converges uniformly to $f$?

Update: After the comment of a user, I thought of adding one additional condition:

  • The sequence $f_v$ is uniformly Lipschitz at $0$, i.e. there is an $L>0$ such that $\|f_v(x)\|\leq L\cdot\|x\|$ for all $x\in \Re^m$ and for all $v\in\mathbb N$.
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Doesn't your additional condition imply equicontinuity of the sequence $(f_v)$? –  David Mitra Aug 2 '12 at 23:17
    
@DavidMitra You're right! I changed it to uniform Lipschitz continuity about the origin. –  Pantelis Sopasakis Aug 2 '12 at 23:22
    
I'm not sure you will be able to weaken the equicontinuity condition. At least in the case where the $f_n$ are functions from some compact metric space $X$ to $\Bbb R$ that converge uniformly to a function $f$, by the Arzela-Ascoli theorem they are necessarily equicontinuous. –  David Mitra Aug 2 '12 at 23:38

1 Answer 1

Let $$f_n(x):=\begin{cases} 2nx&\mbox{ if }0\leq x\leq \frac 1{2n},\\ -2nx+2&\mbox{ if }\frac 1{2n}\leq x\leq \frac 1n\\ 0&\mbox{otherwise}. \end{cases}$$ It's a continuous functions, $|f_n(x)|\leq 1$ for all $x$, converges pointwise to the function equal to $0$ (since for each $x$, we have $f_n(x)=0$ for $n\geq N(x)$), and the limit is obviously continuous. Since $f_n\left(\frac 1{2n}\right)=1,$ we have not the mentioned property, hence we can't have uniform convergence on $[0,1]$.

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Thanks a lot for the answer. Indeed $f_v$ is a very good example. Its graph is like triangles that are squeezed towards the vertical axis! What happens however if we assume that $f_v$ are uniformly Lipschitz (see my update)? –  Pantelis Sopasakis Aug 2 '12 at 23:14

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