# uniform convergence of maxima of function sequence

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.

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.
• 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.