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?

  • $\begingroup$ Looks good. But you should replace ' $\to$ ' (\to) by ' $\implies$ ' (\implies). $\endgroup$ Nov 6 '19 at 8:50
  • 3
    $\begingroup$ 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. $\endgroup$ Nov 6 '19 at 8:51
  • $\begingroup$ @KaboMurphy I agree, thank you $\endgroup$
    – Abhay
    Nov 6 '19 at 8:54
  • $\begingroup$ +1 in spite of an unwarranted assumption. See my A. $\endgroup$ 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 $Y.$

  • $\begingroup$ 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. $\endgroup$ Nov 6 '19 at 10:17

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