I have a sequence of functions of continous functions, say

$f_{n,i} \colon [0,1] \to \mathbb{R}$ with $f_{n,i} \to f_i$ uniformly for each $i \in \mathbb{N}$, i.e. $\sup \limits_x | f_{n,i}(x) - f_i(x) | \to 0$. $f_i$ is continous as well.

Does this imply that $\max \limits_{i \leq n} f_{n,i} \to \max \limits_{i \in \mathbb{N} }f_i$ uniformly?

In particular I would have to show that

$\sup \limits_{x}| \max \limits_{i \leq n} f_{n,i}(x) - \max \limits_{i \in \mathbb{N} }f_i(x) | \to 0$

Here is how far I got:

$\begin{align*} \sup \limits_{x} | \max \limits_{i \leq n} f_{n,i}(x) - \max \limits_{i \in \mathbb{N} }f_i(x) | & = \sup \limits_{x} | \max \limits_{i \leq n} (f_{n,i}(x) - f_i(x)) - \max \limits_{i > n }f_i(x) | \\ & \leq \sup \limits_{x} | \max \limits_{i \leq n} (f_{n,i}(x) - f_i(x)) | + \sup \limits_{x} |\max \limits_{i > n }f_i(x) | \\ & \leq \sup \limits_{x} \max \limits_{i \leq n} | (f_{n,i}(x) - f_i(x)) | + \sup \limits_{x} \max \limits_{i > n } |f_i(x) | \end{align*}$

Here the first term converges to $0$ by the uniform convergence of the $f_{n,i}$. Can I ignore the the second term as taking the limit will run through all the $i \in \mathbb{N}$ anyway?

In fact, by the uniform limit theorem this would prove that taking the maximum of such a sequence $f_{n,i}$ is a continous function.


1 Answer 1


Careful: if no further asumptions are given, you cannot talk about the maximum. For example, the sequence defined by \begin{align} f_n(x) = \left\{\begin{array}{rcl} \dfrac{1}{x} & \text{if} & x >0 \\ \dfrac{1}{n+1} & \text{if} & x = 0 \end{array}\right. \end{align} uniformly converges on $[0,1]$ to the function that is equal to $0$ at $0$ and $1/x$ at $x>0$. But none of them have a maximum value, neither they are bounded.

  • $\begingroup$ By maximum I mean the pointwise maximum. If I have a sequence of functions $f_n$, then I should have a maximum for each $x$. For example: $f_n(x) = x/n$ has a maximum. (Of course, this would always be $f_1(x) $ for $x \in [0,1]$ But you are a totally right, I should have mentioned that the $f_{n,i}$ and the limit $f_i$ are continous. Thank you. I will edit my question. $\endgroup$
    – KiGu
    May 27, 2020 at 14:06
  • $\begingroup$ Oh right, I see. I'll think about that $\endgroup$
    – Didier
    May 27, 2020 at 14:08

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