2
$\begingroup$

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 $n+1$?

$\endgroup$
  • 2
    $\begingroup$ I don't see any reason to use $2n_{k_r}$. Also, I have not seen any proof that requires continuity. $\endgroup$ – Kavi Rama Murthy Apr 22 at 23:22
  • $\begingroup$ Please state the question you are proving rigorously, and avoid "$f$ was not uniformly convergent" this kind of typo. $\endgroup$ – XIAODA QU Apr 23 at 0:28
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.