Suppose there is there is uniform convergent sequence $(f_n)$ on the set $A$, and each $f_n$ is bounded on $A$, i.e., there exist $M_n>0$ such that $|f_n(x)|\le M_n$ for all $x\in A$ Is it true that there exist a function $f$ that is bounded on set A? Can I prove it this way:

Since $f_n \to f$ uniformly on A, $\exists n \ge N$ such that $$|f_N(x) - f(x)| \le 1$$ and $$|f_n(x)-f_N(x)| \le 1$$

Therefore, $|f(x)| \le |f_N(x)|+1 \le M_N + 1$ and $|f_n(x)| \le |f_N(x)| + 1 \le M_N + 1$, $x \in A , n \ge N$

Let $M = \max\{M_1, M_2,\dots,M_N\} + 1$ : $|f_n(x)| \le M$, $\forall x\in A$

Therefore, $f$ is a bounded function on A.

  • 1
    $\begingroup$ What does that colon mean? Also, please don't abuse quantifier notation. $\endgroup$
    – dfeuer
    Oct 13, 2013 at 4:34

1 Answer 1


Your proof is almost correct, but it seems quite awkward at points and doesn't state the correct conditions on variables (you have incorrectly quantified $n$ and $N$, for example). What I think you mean is:

By uniform convergence, there exists an $N$ such that $|f_n(x) - f(x)| \le 1$ for all $n \ge N$. Then $$|f(x)| \le |f_N(x)| + 1 \le M_N + 1$$

Note that we can explicitly state the bound in terms of $M_N$ and don't need to take a maximum to bound $f$ itself.


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