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.

Proof by contradiction:

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 $n+1$?

  • 2
    $\begingroup$ I don't see any reason to use $2n_{k_r}$. Also, I have not seen any proof that requires continuity. $\endgroup$ Apr 22, 2019 at 23:22
  • $\begingroup$ Please state the question you are proving rigorously, and avoid "$f$ was not uniformly convergent" this kind of typo. $\endgroup$
    – Selene
    Apr 23, 2019 at 0:28

1 Answer 1


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 $|f_{B(n)}(x_{A(n)})-f(x_{A(n)})|<r/2$ 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 $B(n)<j\le B(n+1)$ 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.


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