Is the minimum of a sequence of uniformly equicontinuous functions continuous

continuityequicontinuityreal-analysissequence-of-functionuniform-continuity

I have a uniformly equicontinuous (you can even assume that they are uniformly Lipschitz if that helps) sequence of functions $(f_n)_{n=1}^{\infty}$ on $[0,1]$ such that for any $x$, we have that $\epsilon(x):=\min_{n}f_n(x)>0$ (so the minimum is attained for some $n$ and is positive). I want to know if there is some uniform $\epsilon >0$ such that $f_n(x)>\epsilon$ for all $x,n$.

One thing I thought about doing is showing that the function $\epsilon(x)$ is continuous, and then using compactness of $[0,1]$ (and the fact that $\epsilon(x)>0$ pointwise) to show that $\epsilon(x)$ attains some positive minimum which is the $\epsilon$ I want. But I am not sure whether $\epsilon(x)$ is really continuous (intuitively it seems to be, but I know that intuition can be misleading).

I thought about using equicontinuity + some triangle inequality argument in order to claim that:
$$|\epsilon(x)-\epsilon(y)|\leq |f_{n_1}(x)-f_{n_1}(y)|+|f_{n_1}(y)-f_{n_2}(y)|$$
Where $n_1$ is the $n$ for which the minimum in $x$ is attained and $n_2$ is the $n$ for which the minimum in $y$ is attained. The first term is 'small' from equicontinuity, and the second term should intuitively be small since $f_{n_1}$ and $f_{n_2}$ are both 'close' to the minimum, but I am not sure on how to explain the second part (if it is true).

I'd appreciate any help (both in showing continuity or in solving the original problem). Of course, counter examples are also welcome 🙂

Thanks in advance.

Best Answer

By definition of uniformly equicontinuity, for any $\delta >0$ there is $\eta>0$ so that $$|f_n(x_1) - f_n(x_2)|<\delta$$ for all $n$ and for all $x_1, x_2\in [0,1]$ with $|x_1-x_2|<\eta$.

Let $x_1, x_2\in [0,1]$ with $|x_1- x_2|<\eta$. Let $f_{n_1}$ be chosen ($n_1$ depends on $x_1$) so that $$ f_{n_1}(x_1)=\epsilon (x_1).$$

Then $$\epsilon (x_2) - \epsilon (x_1) \le f_{n_1}(x_2) - f_{n_1} (x_1)\le |f_{n_1}(x_2) - f_{n_1} (x_1)|<\delta. $$

Similarly $\epsilon(x_1) - \epsilon(x_2) < \delta$ (by interchanging the role of $x_1, x_2$ in the above argument).

Thus we have

$$|\epsilon (x_1) - \epsilon (x_2)|<\delta$$ whenever $x_1, x_2\in [0,1]$ and $|x_1-x_2|<\eta$. Thus $\epsilon(x)$ is continuous.

Remark

  • Indeed we showed that $\epsilon$ is uniform continuous.
  • The domain can be changed to any metric space and the above still works.
  • Similar argument also shows that $g:= \inf_n f_n$ is uniform continuous (if $g$ is assumed finite everywhere; see here for example).
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