# Sequence of elements having a convergent subsequence -NBHM $2014$

Question is to find which of the following are true?

1. Let $V$ be the space of continuous functions on $\mathbb{R}$ with compact support endowed with metric $$d(f,g)=\bigg(\int_{-\infty}^{\infty}|f(t)-g(t)|^2\bigg)^{\frac{1}{2}}$$Let $f:\mathbb{R}\rightarrow \mathbb{R}$ which vanish outside $[0,1]$. define $f_n(x)=f(x-n)$ for $x\in \mathbb{N}$. Then $(f_n)$ has a convergent sub sequence in $V$.
2. Let $\varphi, \psi$ be continuous function on $[0,1]$. Let $(f_n)$ be a sequence in $\mathcal{C}[0,1]$ with its sup norm topology suchthat, for all $n\in \mathbb{N}$ the functions $f_n$ are continuously diffrerentiable and for all $x\in[0,1]$ and for all $n\in \mathbb{N}$ we have $|f_n(x)|\leq \varphi(x)$ and $|f_n'(x)|\leq \psi(x)$ Then there exists a sub sequence of $(f_n)$ which converges in $\mathcal{C}[0,1]$.
3. Let $\{A_n\}$ be a sequence of orthogonal matrices in $M_n(\mathbb{R})$. then it has a convergent subsequence.

I am sure that $3$ is true as Orthogonal matrices are compact and every sequence in compact space has a convergent subsequence.

I see that $2$ is also correct by Arzela Ascoli Theorem.

I am not so sure about $1$ and actually i did not understand the idea behind that.

Thank you.

• Continuous function with compact support are dense in $C[0,1]$ – Marso Jan 27 '14 at 10:54
• @Tojamaru So.. how does that help me in this case? – user87543 Jan 27 '14 at 10:54
• For (1): if exists the convergent subsequence, the limit is... – Martín-Blas Pérez Pinilla Feb 12 '14 at 8:46
• @Martín-BlasPérezPinilla : I am sorry, i could not get your point... – user87543 Feb 12 '14 at 8:49
• What's the meaning of "with compact support endowed"? – Hoseyn Heydari Feb 12 '14 at 9:02

For any supposed limit $g$ (continuous function with compact support), take $n_0$ large enough s.t. for $n\ge n_0$: ${\rm supp}\,f_n\cap{\rm supp}\,g=\emptyset$. Now, if $n_k\ge n_0$: $$d(f_{n_k},g)^2=\int_{-\infty}^{\infty}|f_{n_k}(t)-g(t)|^2dt= \int_{-\infty}^{\infty}|f_{n_k}(t)|^2dt+\int_{-\infty}^{\infty}|g(t)|^2dt\ge$$
$$\ge\int_{-\infty}^{\infty}|f_{n_k}(t)|^2dt =\int_{-\infty}^{\infty}|f(t)|^2dt,$$ and this implies $f=0$. The property is only true in this trivial case.
• I am sorry, I did not get your point.... why would ${\rm supp}\,f_n\cap{\rm supp}\,g=\emptyset$ and why would last equality hold :O – user87543 Feb 12 '14 at 9:57
• Draw some $f$. Draw $f_n$ (it will be a translated of $f$) for several $n$. – Martín-Blas Pérez Pinilla Feb 12 '14 at 10:00
• Excuse me for my dumbness but i am unable to think of a continuous function which vanish outside $[0,1]$.. I can think of continuous functions on $[0,1]$ but they are not zero outside $[0,1]$.. No obvious example is coming to my mind... – user87543 Feb 12 '14 at 11:22