1
$\begingroup$

Given a sequence of functions $f_n: (X,d_X) \rightarrow (Y,d_Y)$ where each function $f_n$ is bounded, I want show that if the $f_n$ converge uniform to some function $f:X \rightarrow Y$ then $f$ has to be bounded.

This seems quite obvious but I am not sure how to approach this. Does proof by contradiction work best ?

  • 3
    The question is subjective, but I would say no, direct proof is better here. There's an $n$ such that $d_Y(f_n(x),f(x))<1$ for all $x$, and you can use boundedness of $f_n$ and the triangle inequality.2012-12-03
  • 1
    Yes indeed. Let $y \in Y$. Then there are $R_n$ such that $\forall n \in \mathbb N: f(X) \subseteq B(y,R_n)$. Let then $N \in \mathbb N$ such that for all $x \in X: d_Y(f_N(x),f(x)) < 1$. Then we get for all $x \in X$ that $d_Y(f(x),y) \leq d_Y(f(x),f_N(x))+d_Y(f_N(x),y) < 1 + R_N$ and thus $f(X) \subseteq B(y,1+R_N)$.2012-12-03

0 Answers 0