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Prove directly that if $f_n$ converges uniformly to $f$ and if $f_n$ converges pointwise to $g$, $f=g$.

Thoughts/Attempts at a Solution: I know that if $f_n$ converges uniformly to f , then by definition for every $e> 0$, there exists an N so that $n\geq N$ implies that $|f_n(x)-f(x)|<e$ holds for all $x\in E$.

If $f_n$ converges pointwise to g, then by definition, for every $x \in E$ and for every $e> 0$, there exists an N so that for every for $n\geq N$ we have $|f_n(x)-g(x)|<e$.

I know that the basic difference is that in point wise convergence N can depend on both x and $e$ as both are quantified first while in uniform convergence we need an N that solely depends on $e$. However, now I don't know how to use these two definitions to explicitly show that f=g.

Usually when I do proofs regarding uniqueness, I do a proof by contradiction, but in this case I have to do a direct proof so I'm not exactly sure how to proceed. Any help would be much appreciated.

kemb
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Hint: uniform convergence implies pointwise convergence, in the definition of uniform convergence if $N$ is found for all $x$, and $|f_n(x)-f(x)|<e$ it implies that for a fixed $x_0$, $|f_n(x_0)-f(x_0)|<e$, for $n>N$ so if $f_n$ converges uniformly to $f$ it implies that $f$ converges pointwise to $f$ and the limit is unique.

Tsemo Aristide
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  • Thanks. I know uniform convergence implies pointwise convergence. My only problem is that I don't know how to show the limit is unique. – kemb Feb 08 '18 at 10:15
  • Is there a way to explicitly show f(x) must equal g(x)? Thanks. – kemb Feb 08 '18 at 10:15
  • https://math.stackexchange.com/questions/882250/proving-that-a-convergent-sequence-has-a-unique-limit – Tsemo Aristide Feb 08 '18 at 10:16
  • Thanks. Can I write something up and verify with you that it is correct? Thanks. – kemb Feb 08 '18 at 10:19
  • For every $e> 0$, there exists an N so that $n\geq N$ implies that $|f_n(x)-f(x)|<e/2$ holds for all $x\in E$ and (furthermore since we have uniform convergence we have point wise convergence) $|f_n(x)-g(x)|<e/2$ . We thus have for n>N, $|f(x)-g(x)| <= |f(x)-f_n(x)|+|f_n(x)-g(x)|<e/2+e/2=e$. Hence f(x)=g(x). Is this correct? – kemb Feb 08 '18 at 10:26
  • yes it is correct. – Tsemo Aristide Feb 08 '18 at 12:13