# If $(f_n)$ converges uniformly to $f$, then prove that $\lim_{x\to c}\lim_{n\to \infty} f_n(x) = \lim_{n\to \infty} \lim_{x\to c}f_n(x)$

Let $$(f_n)$$ be sequence of real valued functions on $$X\subseteq R$$. If $$(f_n)$$ converges uniformly to $$f$$, then prove that $$\lim_{x\to c}\lim_{n\to \infty} f_n(x) = \lim_{n\to \infty} \lim_{x\to c}f_n(x)$$

Since, $$(f_n)$$ converges uniformly to $$f$$, $$(f_n)$$ is uniformly Cauchy and hence,

for $$\epsilon>0\; \exists N\in \mathbb{N}$$ such that $$|f_n(x) - f_m(x)|<\epsilon \; \forall m,n\geq N$$ and $$x\in X$$ Hence,

$$\lim_{x\to c} |f_n(x) - f_m(x)|\leq \epsilon \; \forall m,n\geq N, x\in X$$

Let $$\lim_{x\to c}f_n(x) = A_n$$,

then we have, $$|A_n - A_m|<\epsilon \; \forall m,n\geq N$$

So, $$(A_n)$$ is Cauchy in $$R$$ and hence converges to some $$A_0\in R$$

I want to show $$\lim_{x\to c} f(x) = A_0$$ so that I will have $$\lim_{x\to c}\lim_{n\to \infty} f_n(x) = \lim_{x\to c}f(x)= A_0 = \lim_{n\to \infty}A_n = \lim_{n\to \infty} \lim_{x\to c}f_n(x)$$

Now, as $$\lim_{x\to c}f_n(x) = A_n \rightarrow \; \forall \epsilon >0 \; \exists \delta>0$$ such that if $$x\in X, \;0<|x-c|<\delta \rightarrow |f_n(x)-A_n|<\frac\epsilon 3 \; \forall n\in N$$

Since $$A_n \to A_0 \rightarrow \exists M\in N$$ such that $$|A_n - A_0|<\frac\epsilon 3 \; \forall n\geq M$$

Also since $$(f_n) \to f$$ uniformly on $$X$$, $$\exists J\in N$$ such that $$|f_n(x)-f(x)|<\frac\epsilon 3 \forall n\geq J$$ and $$\forall x\in X$$

Finally letting $$T=max{M,J}$$ and letting $$x\in X, \; 0<|x-c|<\delta$$ we have,

$$|f(x)-A_0|\leq |f(x)-f_T(x)|+|f_T(x)-A_T|+ |A_T - A_0| < \epsilon$$

This proves it.

Is this correct?

• Looks good. But you should replace ' $\to$ ' (\to) by ' $\implies$ ' (\implies). Nov 6 '19 at 8:50
• Minor correction: A strict inequality doesn't remain strict when you take limits. For example $1-\frac 1 n <1$ for all $n$ but $1 <1$ is false. Nov 6 '19 at 8:51
• @KaboMurphy I agree, thank you Nov 6 '19 at 8:54
• +1 in spite of an unwarranted assumption. See my A. Nov 6 '19 at 9:51

It is correct if you assume that $$f_n$$ is continuous at $$c$$ for all but finitely many $$n.$$ Otherwise the RHS may fail to exist.

The 6th line of your Q (the def'n of $$A_n$$) is that assumption.

E.g. let $$X=\Bbb R,\;$$ let $$\chi_{\Bbb Q}$$ be the characteristic (indicator) function of $$\Bbb Q,$$ i.e. $$\chi_{\Bbb Q}(x)$$ is $$1$$ if $$x\in \Bbb Q\,;$$ otherwise it is $$0.$$ Let $$f_n(x)=\chi_{\Bbb Q}(x)/n.$$ Then $$f_n$$ converges uniformly to $$0.$$

So the LHS is always $$0$$.

But for any $$c\in \Bbb R$$ the limit $$\lim_{x\to c}f_n(x)$$ does not exist for $$any$$ $$n,$$ so $$\lim_{n\to \infty}\lim_{x\to c}f_n(x)$$ doesn't exist either.

The most common application is that if each $$f_n$$ is continuous at all $$c\in X$$ then $$f$$ is, also. And the proof applies, almost verbatim, to the uniform limit of a sequence of continuous functions from any metric space $$X$$ to any metric space $$Y.$$

• An important application: Let $r>0$ and $D=\{z\in \Bbb C: |z|<r\}.$ Let $f_n(z)=\sum_{j=0}^nA_jz^j.$ We can (fairly easily) show that if $(f_n(z))_n$ converges to $f(z)$ for each $z\in D,$ then both $(f_n)_n$ and $(f'_n)_n$ converge uniformly on any closed $C\subset D$. So with $g(z)=\lim_{n\to \infty}f'_n(z),$ we have $g$ continuous on $D$. Now (fairly easily) we show that $f'=g.$ That is, the power series $f(z)=\sum_{j=0}^{\infty}A_jz^j$ is analytic and can be differentiated term-by-term on D. Nov 6 '19 at 10:17