Here is my problem and my solution appear subsequently. please check to see if there are faults

Consider the piecewise linear functions in $C[-1,1]$ given by $f_n(x)=0$ for $-1\leq x\leq 0$, $f_n(x)=nx$ for $0\leq x\leq \frac1n$, and $f_n(x)=1$ for $\frac1n\leq x\leq 1$ with uniform norm($\Vert f\Vert_\infty=\sup_{x\in K} \vert f(x)\vert$)

(a) show that $\Vert f_n-f_m\Vert_\infty \ge \frac{1}{2}$ if $m\ge 2n$

(b) hence show that no subsequence of $(f_n)_n^\infty$ converges

I can show above problem (a). so suppose (a) is true. here is my solution about (b). I want to make sure whether it is right or not.

pf) suppose to the contrary that there is convergent subsequence of $(f_n)_n^\infty$.

let$$ \lim_{k \to\infty}f_{n_k}=f$$ and we already know that every convergent seuqence is a Cauchy sequence. So $(f_{n_k})_{k=1}^\infty$ is also cauchy then we choose $m\ge 2n_k$ like above (a) $$\Vert f_{n_k}-f_m\Vert_\infty\ge \frac{1}{2}$$ it contradicts the definition of a Cauchy sequence thus every subsequence of $(f_n)_1^\infty$ is not convergent.

addtionally, I have a general problem on solving this.

"every cauchy sequence that isn't convergent to some vector in normed vector space has no convergent subsequence?"

  • $\begingroup$ Having a Cauchy subsequence does not imply the sequence itself is Cauchy. $\endgroup$ – lzralbu Sep 3 '17 at 13:00
  • $\begingroup$ I replace" $(f_n)_{n=1}^\infty$ is cauchy" with $(f_{n_k})_{k=1}^\infty$ is cauchy" $\endgroup$ – fivestar Sep 3 '17 at 13:07

You edited proof is pretty much correct:

Since $(f_{n_k})_{k=1}^\infty$ is convergent, you can conclude that is is also Cauchy.

However, for any $l\in\mathbb{N}$ find $k>l$ such that $n_k \geq 2n_l$ so we have $\|f_{n_k} - f_{n_l}\|_\infty \geq \frac{1}{2}$. This is a contradiction with the definition of Cauchyness since for $\varepsilon = \frac{1}{2}$ you cannot find $n_0 \in \mathbb{N}$ such that $k, l\geq n_0 \implies \|f_{n_k} - f_{n_l}\|_\infty < \frac{1}{2}$.

Thus, $(f_{n_k})_{k=1}^\infty$ does not converge.

There is another way to prove that no subsequence of $(f_n)_{n=1}^\infty$ converges in the $\|\cdot\|_\infty$ norm, without using (a):

Notice that $(f_n)_{n=1}^\infty$ converges pointwise to the function $\chi_{(0,1]}$. Then, every subsequence $(f_{n_k})_{k=1}^\infty$ also converges pointwise to $\chi_{(0,1]}$. Since uniform convergence implies poinwise convergence, the only candidate for the uniform limit of $(f_{n_k})_{k=1}^\infty$ is $\chi_{(0,1]}$, which is not a continuous function and thus not in $C([-1,1])$.

About your last remark, this is equivalent to the fact that if a subsequence of a Cauchy sequence converges, then the whole sequence converges (to the same limit as the subsequence):

Let $(x_n)_{n=1}^\infty$ be a Cauchy series in a normed space $(X, \|\cdot\|)$, such that the subsequence $(x_{p(n)})_{n=1}^\infty$ converges to the limit $x \in X$.

Take $\varepsilon > 0$. There exists $n_0 \in \mathbb{N}$ such that $n \geq n_0 \implies \|x_{p(n)} - x\|<\frac{\varepsilon}{2}$. Also, there exists $n_0 \in \mathbb{N}$ such that $m, n \geq n_1 \implies \|x_m - x_n\|<\frac{\varepsilon}{2}$. For $n \geq \max\{n_0, n_1\}$ we have: $$\|x_n - x\| \leq \|x_n - x_{p(n)}\| + \|x_{p(n)} - x\| < \varepsilon.$$ Thus, $\lim_{n\to\infty} x_n = x$.

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  • $\begingroup$ @merchanodroid I always appreciate your interest. thank you very much! I am wrong about typing interval. It is not $m\ge n$ but $m \ge 2n$. so if the interval is before I wrote. I understand your solution is true. and I read your solution of my first question well. it help me understand better. $\endgroup$ – fivestar Sep 3 '17 at 13:23
  • $\begingroup$ @merchanodrod I have not studied yet about pointwise convergence and uniform convegence of functional sequence. but I quess uniform convergence is equivalent to the pointwise convergence. and unform convergence implies $\chi_\left(0.1\right]$ must be continuous function. but it is contradiction. right? and I understand your explanation of my last remark. thanks a lot! $\endgroup$ – fivestar Sep 3 '17 at 15:09
  • $\begingroup$ Uniform convergence is convergence in the uniform norm $\|\cdot\|_\infty$, and pointwise convergence of $(f_n)_{n=1}^\infty$ to the function $f$ is simply $\lim_{n\to\infty}f_n(x) = f(x)$, $\forall x \in [-1, 1]$. It is easy to see that uniform convergence implies pointwse convergence since for any $x \in [-1, 1]$ we have $|f_n(x)-f(x)|\leq \|f_n-f\|_\infty$. The converse is not true in general, your sequence is an example. $\endgroup$ – mechanodroid Sep 3 '17 at 16:06
  • $\begingroup$ thank you ! I understand what you said.! $\endgroup$ – fivestar Sep 4 '17 at 10:00

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