# Pointwise and Uniform Convergence. Showing unique limit.

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.

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.
• 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 8 '18 at 10:26