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Suppose $f, f_{1,\cdots,n}\in X^*$ where $X$ is a normed vector space. How to show that, if $\cap Ker f_i \subset Ker f$, then $f$ is a linear combination of $f_{1,\cdots,n}$?

I'm at a loss even for the classical linear algebra cases (where $X$ is a finite dimensional Euclidean spade with $\ell^2$ norm, so really don't know how to proceed for the more general cases.

EDIT It seems I might want to relax the conditions a little here. So let's now assume $X$ is a Banach space.

EDIT The Banach space assumption can be dropped. Any normed vector space is okay.
Vim
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2 Answers2

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Another proof by my teacher: define $E=\{(f_1(x),\cdots,f_n(x)), x\in X\}$, a subspace of $K^n$ ($K$ denotes the real or complex field), and define $g:E\to K$ as $g(f_1(x),\cdots,f_n(x)):=f(x)$ (well defined by the kernel inclusion). Now we can extend $g$ to $g':K^n\to K$ since $E$ being a finite dimensional subspace is closed, and then $$f(x)=g(f_1(x)e_1+\cdots+f_n(x)e_n)=g'(f_1(x)e_1+\cdots+f_n(x)e_n)=\sum f_i(x)g'(e_i).$$

Vim
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Notice that we can reduce the problem to a finite dimensional $X$ by dividing out the closed subspace $K=\bigcap_{i=1}^n \ker f_i$.

Let $\mathbb F$ be the field of real or complex numbers.

  1. Let $\bar f_1, \dotsc, \bar f_n, \bar f : X/K \to \mathbb C$ denote $f_1, \dotsc, f_n, f$ modulo $K$, that is $$ \bar f_i(x+K) = f_i(x) $$ for every $x\in X$ and so on. The linear map $g:X/K \to \mathbb F^n$ defined by $g_i = \bar f_i$ has trivial null space and hence is injective. Thus, $X/K$ must be finite dimensional. Assume there is $\alpha\in \mathbb F^n$ with $$ \sum_{i=1}^n \alpha_i \bar f_i = \bar f. $$ Then, for the same $\alpha$ we have $$ \sum_{i=1}^n \alpha_i f_i = f. $$

  2. Now the proof for the finite dimensional case:

    Let $F = span\{ f_1, \dotsc, f_n \}$. Then, $F$ is closed. Assume the converse, that is $f\notin F$. By hahn banach separation theorem there is a $\lambda\in X^{**}$ (isomorph to $X$) such that $$ Re\lambda(f) < \inf_{\alpha\in\mathbb F^n} Re\lambda\left(\sum_{i=1}^n \alpha_i f_i \right) = \inf_{\alpha\in\mathbb F^n} \sum_{i=1}^n Re(\alpha_i \lambda(f_i)). $$ That implies $\lambda(f_i) = 0$ for every $i$ (otherwise consider $\alpha_{i,n} = -n\overline{\lambda(f_i)}$). That is $$ \lambda\in \bigcap_{i=1}^n \ker f_i \subseteq \ker f. $$ Clearly a contradiction.

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user251257
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  • Thanks. But two things I don't understand: 1). How is it implied that $\langle \lambda,f_i\rangle=0$, and 2). What's the contradiction here? I guess we have to create a $\lambda\in Ker f$ but not in $\cap Ker f_i$. – Vim Nov 05 '16 at 03:27
  • @Vim: 1.) If say $\langle \lambda, f_1 \rangle > 0$ consider $\alpha_1 \to -\infty$. 2.) we have $0 = \langle \lambda, f \rangle < \inf_{\alpha} \sum_i \alpha_i 0 = 0$. – user251257 Nov 05 '16 at 03:29
  • thanks. Another follow up one: what version of Hahn Banach were you using here? I can't identify it with those I learned. – Vim Nov 05 '16 at 03:34
  • Ok. I now fully understand your proof and see the reason why you require the space be real and self-reflexive. However, since it is more natural to require a space to be Banach than to be self-reflexive etc., would you offer any insights into the Banach space case, if you please? – Vim Nov 05 '16 at 03:48
  • I guess your $g$ is a vector valued functional? – Vim Nov 05 '16 at 04:07
  • never mind let me take some time to digest it. – Vim Nov 05 '16 at 04:10
  • @user251257 In fact, there is no need of topology. The statement holds (exactly with your proof) for any vector space (finite or infinite dimensional and over any field $k$) with a separation of $f$ and $F$ by a hyperplane and replacing "inf" by "for all $\alpha\in k^n$". – Duchamp Gérard H. E. Nov 05 '16 at 04:21
  • @user251257 Of course not, but you first factor, as you did, by the intersection of the kernels. – Duchamp Gérard H. E. Nov 05 '16 at 04:35
  • Given $X$ isomorphic to its double dual, how do you relate $\lambda(f_i)$ with $f_i(\lambda)$? (What if $X$ isn't a Hilbert space) – Vim Nov 05 '16 at 05:15
  • @DuchampGérardH.E. does the separation theorem hold for any vector spaces? I guess at least the underlying field must be "dense" ($\Bbb C$ or $\Bbb R$) and there might be something else required. – Vim Nov 05 '16 at 05:20
  • @Vim It does in every finite dimensional space, this is the situation reached after factoring by the intersection of kernels (see user251257's proof replacing $\mathbb{C}$ by $k$). – Duchamp Gérard H. E. Nov 05 '16 at 05:26
  • @user251257 doesn't seem to be a problem. Just replace $\lambda$ with $\Re \lambda$? – Vim Nov 05 '16 at 05:41
  • Also I just got another proof. I've added it as an answer. – Vim Nov 05 '16 at 05:50
  • @Vim I fixed it. – user251257 Nov 05 '16 at 05:53