# If $x_n \rightarrow x \implies \lim_{n \rightarrow +\infty}$ $f_n (x_n) \rightarrow f(x)$ ,then $f_n \rightarrow f$ uniformly in each compact set $K$

I found this proof online and was having trouble understanding the last steps.

Lemma: Given that $$f_n(x_n)$$ converges to $$f(x)$$ , show that $$f_n$$ is uniformly convergent to $$f$$ on a compact set k.

Assume $$f_n$$ (on compact set K) does not converge uniformly to $$f$$. Then there exists a sequence of $$x$$'s and an increasing sequence of $$n \epsilon N$$ s.t. $$| \; f_{2n_{k_r-1}}(x_k) \; - \; f_{2n_{k_r}}(x _k) \; | \geq \epsilon .$$

Then, by Bolzano Weierstrass (since K is compact) there is a convergent subsequence {$$x_{k_r}$$} that converges to $$x$$.

Define a sequence $$y_{n}$$ so that

$$y_{n_r} = x\; \textrm{if} \; n \neq 2n_{k_r} \; \textrm{or if} \; n \neq 2n_{k_r-1},$$

and $$y_{2n_{k_r}} = y_{2n_{k_r-1}} = x_{k_r} .$$ Then $$y_n$$ conveges to $$x$$.

let {$$Z_n$$} = {$$f_n(y_n)$$} . This sequence is not cauchy since $$| \; z_{2n_{k_r-1}} \; - \; z_{2n_{k_r}} \; | \geq \epsilon.$$

Thus we have proved that if $$f_n$$ does not converge uniformly to $$f$$,

$$x_n \rightarrow x \implies \lim_{n \rightarrow +\infty}$$ $$f_n (x_n) \rightarrow f(x)$$ is false

My question is, couldn't we just use {$$Z_n$$} = {$$f_n(x_{k_r}))$$}? Why did they define a new sequence that is potentially just {$$x_{k_r}$$}?

also, is all this $$2n-1$$, $$2n$$ notation necessary? could we use $$n$$ and $$n+1$$?

• I don't see any reason to use $2n_{k_r}$. Also, I have not seen any proof that requires continuity. – Kavi Rama Murthy Apr 22 '19 at 23:22
• Please state the question you are proving rigorously, and avoid "$f$ was not uniformly convergent" this kind of typo. – Selene Apr 23 '19 at 0:28

If $$f_n$$ does not converge uniformly on some compact $$K$$:

There exists $$r>0$$ and a sub-sequence $$(A(n))_n$$ of $$\Bbb N$$ and a sequence $$(x_{A(n)})_n$$ in $$K$$, converging to $$x\in K,$$ such that $$|f_{A(n)}(x_{A(n)})-f(x_{A(n)})|>r$$ for all $$n.$$

Now there exists a sub-sequence $$(B(n))_n$$ of $$\Bbb N$$ such that $$B(n)>A(n)$$ and $$|f_{B(n)}(x_{A(n)})-f(x_{A(n)})| for all $$n.$$

We would like to have a sequence $$Y=(y_n)_n$$ in $$K,$$ converging to $$x,$$ whose $$B(n)$$-th entry $$y_{B(n)}$$ is $$x_{A(n)}$$ for every $$n$$. We would then have $$f(x)=\lim_{n\to \infty}f_{B(n)}(y_{B(n)})=\lim_{n\to \infty}f_{B(n)}(x_{A(n)}).$$

Then the sequences $$S_1=(f_{A(n)}(x_{A(n)})_n$$ and $$S_2=(f_{B(n)}(y_{B(n)}))_n=(f_{B(n)}(x_{A(n)})_n$$ both converge to $$f(x).$$ But for every $$i$$, the $$i$$-th term of $$S_1$$ and the $$i$$-th term of $$S_2$$ differ by more than $$r/2,$$ which makes it impossible for both series to converge to $$f(x).$$ So we obtain the desired paradox.

To obtain $$Y$$ let $$y_j=x_{A(1)}$$ for $$j\le B(1),$$ and for $$B(n) let $$y_j=x_{A(n+1)}.$$

Remarks. (1). We could combine $$S_1$$ and $$S_2$$ into a single sequence before deriving the contradiction, which seems to be the style of the proof you found, but it's not necessary. (2). The converse to this result is also true.