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From Wikipedia

Given a topological space $X$, we can equip the space of bounded real or complex-valued functions over $X$ with the uniform norm topology. Then uniform convergence simply means convergence in the uniform norm topology.

I was wondering

  1. Is convergence in the uniform norm topology same as convergence wrt the uniform norm?
  2. what relations are between convergence wrt the uniform norm, and uniform convergence? Contrary to what Wiki says, I think that convergence wrt the uniform norm implies uniform convergence, but not vice versa. This is because the uniform norm is defined to be taking sup over the domain, stronger than the case of uniform convergence where "for each distance in codomain, there exists a distance in domain and consider all pair of domain points apart within this distance". Am I wrong?

Thanks and regards!

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up vote 3 down vote accepted

Yes, you are wrong: they are exactly the same. If $f_n$ converges uniformly to $g$ on $X$, for every $\epsilon > 0$ there is $N$ such that whenever $n > N$, $|f_n(x) - g(x)| < \epsilon$ for all $x\in X$. But that says $\sup_{x \in X} |f_n(x) - g(x)| \le \epsilon$. The difference between $\le$ and $<$ is not significant: if you want $\sup_{x \in X} |f_n(x) - g(x)| < \epsilon$, just take an $N$ that implies $|f_n(x) - g(x)| < \epsilon/2$.

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The topology is give by the supremum norm $\| f\|_\infty=sup\{|f(x)|: x\in X\} $, and sequence converges $ f_n$ converges to $f$ iff $f_n$ converges uniformly to $f$.

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