Let $X$ be the Banach space of the real valued continuous function in $[0,1]$ endowed with the supremum norm. Suppose $E \subset X$ is a finite dimensional subspace and take $\{f_n\}_{n \in \mathbb{N}} \subset E$ s.t. $$ \lim_{n \to + \infty} \|f_n\|= + \infty $$
I want to prove that there exist a segment $\Delta$ (a connected closed subset of $[0,1]$) and a subsequence $\{f_{n_m}\}_{m \in \mathbb{N}} \subset \{f_n\}_{n \in \mathbb{N}}$ s.t. $$ \lim_{m \to + \infty} \min_{x \in \Delta} |f_{n_m}(x)| = + \infty$$
This is what I have tried:
I can assume wlog to extract subsequences and obtain that
- Call $\{x_n\}$ the sequence of points where the maximum of the $f_n$ are achieved and $x$ a point to which they converge.
- I can decompose each $f_n(x) = \sum_{k=1}^N a_k^n e_k(x)$ with all the coefficients $a_k^n$ eventually positive and $\{e_1, \dots, e_N \}$ a basis for $E$.
- Fix some $\epsilon_k>0$ (we will choose them later) s.t. there exists a closed interval $\Delta \subset [0,1]$ s.t. $$|e_k(x)-e_k(y)| < \epsilon_k$$ for all $x,y \in \Delta$ and $k=1, \dots, N$.
- Eventually the sequence $\{x_n\}$ will be contained in $\Delta$ and then I am able to say that $$e_k(x)> e_k(x_n) - \epsilon_k$$ for each $x \in \Delta$.
- Finally I can estimate for $x \in \Delta$ that $$f_n(x) > f_n(x_n) -\sum_{k=1}^N a_k^n \epsilon_k$$
- I would like to choose the $\epsilon_k$ in order to get divergence but I do not know how to do.
Using Banach Steinhaus theorem I know that there exist some $G_{\delta}$ dense subset of $[0,1]$, call it $G$, s.t. $$\sup_{n \in \mathbb{N}} {f_n(x)}= + \infty \quad \quad \text{ for all } x \in G $$. Can I use this in some way?
Any hint will be very appreciated!
If $E$ is a finite-dimensional vector subspace of a normed linear space then
(1) $E$ is closed,
(2) A closed bounded subset of $E$ is compact, and a bounded sequence in $E$ has a sub-sequence converging to a member of $E.$
There exists some $k_0\le N$ and a sub-sequence $S_0=(n_j)_{j\in \Bbb N}$ of $\Bbb N$ such that $$\forall j\in \Bbb N\;(j<|a_{k_0}^{n_j}|=\max \{|a_k^{n_j}|: k=1,...,n\}).$$
Let $g_{n_j}=(a_0^{n_j})^{-1}f_{n_j}.$
Then $(g_{n_j})_{j\in \Bbb N}$ is a bounded sequence in $E$ so by (2) there is a sub-sequence $S_1=(m_j)_{j\in \Bbb N}$ of $S_0$ and a $g\in E$ such that $$\lim_{j\to \infty}\|g-g_{m_j}\|=0.$$
Now if $N>1$ let $E'$ be the vector subspace generated by $\{e_k:1\le k\le N\}\setminus \{e_{k_0}\},$ or if $N=1$ let $E'$ be the ($0$-dimensional) vector subspace $\{0\} .$ By (1) applied to $E',$ the distance $r$ from $e_{k_0}$ to $E'$ is positive , so $$0<r\le\inf \{\|g_{m_j}\|: j\in \Bbb N\}.$$ So for each $j$ let $x_{m_j}\in [0,1]$ such that $|g_{m_j}(x_{m_j})|\ge r.$
Let $S_2=(p_j)_{j\in \Bbb N}$ be a sub-sequence of $S_1$ and let $x\in [0,1]$ such that $\lim_{j\to \infty}x_{p_j}=x.$
Now $g_{p_j}$ converges uniformly on $[0,1]$ to $g,$ and $x_{p_j}$ converges to $x,$ and $g$ is continuous, so $g_{p_j}(x_{p_j})$ converges to $g(x).$ And since each $|g_{p_j}(x_{p_j})|\ge r$ we have $$|g(x)|\ge r. $$
Let $D$ be a closed interval of positive length with $x\in D\subset [0,1]$ such that $\forall y\in D\,(|g(y)-g(x)|<r/3).$
Let $j_0\in \Bbb N$ such that $j\ge j_0\implies \|g-g_{p_j}\|<r/3.$
Note that for $j\ge j_0$ we have $p_j\ge m_j\ge n_j\ge j$ so $$|a_{k_0}^{p_j}|>p_j\ge j.$$
Now for all $y\in D$ and all $j\ge j_0$ we have $$|g_{p_j}(y)|\ge |g(y)|-\|g-g_{p_j}\|>$$ $$>|g(y)|-r/3\ge$$ $$\ge |g(x)|-|g(x)-g(y)|-r/3>$$ $$>|g(x)|-r/3-r/3\ge$$ $$\ge r-r/3-r/3=r/3.$$
Finally for all $y\in D$ and all $j\ge j_0$ we have $$|f_{p_j}(y)|=|a_{k_0}^{p_j}|\cdot |g_{p_j}(y)|>j\cdot r/3.$$