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Let $X$ be a Hilbert space and $\{F,F_1,\ldots,F_N\}$ linear functionals over $X$ such that $$\bigcap_{i=1}^N\mbox{ker}(F_i)\ \subseteq \mbox{ker}(F),$$ and any kernel of the involved functionals is not dense on $X$. Using the orthogonality (not Hahn-Banach) in order to prove that there exists scalars $\alpha_1,\alpha_2,\ldots,\alpha_N$ such that $$F\ =\ \sum_{i=1}^N\alpha_iF_i.$$


This is the unanswered last part of the question

Hahn-Banach theorem (second geometric form) exercise

Thanks in advance.

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2 Answers 2

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In a Hilbert space $X$ every (bounded) linear functional $F$ is of the form $\langle x,-\rangle$ for some $x\in X$, and thus $\ker F=x^\perp$.

(Because if $F\ne 0$, there is an $x_0$ such that $F(x_0)=\|F\|^2$, and project it orthogonally to the (closed subspace) $\ker F$.)

Then we can write $F_i=\langle x_i,-\rangle$ and we have $$\bigcap_i\,\ker F_i={\rm span}(x_1,..,x_N)^\perp$$ so $x^\perp\supseteq {\rm span}(x_1,..,x_N)^\perp$, therefore $x\in {\rm span}(x_1,..,x_N)$.

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    $\begingroup$ F is not necessary bounded $\endgroup$
    – FASCH
    May 2, 2013 at 0:04
  • $\begingroup$ Then I don't know how to use orthogonalities.. $\endgroup$
    – Berci
    May 2, 2013 at 7:46
  • $\begingroup$ I could prove that $F$ is bounded, using orthogonality. $\endgroup$
    – FASCH
    May 2, 2013 at 17:32
  • $\begingroup$ How? $\ker F$ should be closed, but why it would be? $\endgroup$
    – Berci
    May 2, 2013 at 22:59
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    $\begingroup$ $F$ is linear, then you want to prove that is bounded. Now, $\overline{\mbox{ker} F} \neq X$, then $\overline{\mbox{ker} F}$ is closed proper subspace of a Hilbert $X$. Next, exists $u\neq 0$, that $u\in \overline{\mbox{ker} F}^{\perp}$, and follow the proof of Riesz Representation Theorem you can show that $F(x) = <x, \frac{\overline{F(u)}}{\|u\|}u>$, for all $x\in X$. Then $F$ es bounded. $\endgroup$
    – FASCH
    May 3, 2013 at 3:35
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First of all, note that you can assume $F_i$ to be linearly independent. If not, just argue that removing from the list a linear combination of other functionals does not change things.

Next, show that you can find $x_i \in X$ such that $F_i(x_j) = \delta_{ij}$. This can be done by induction on the number of functionals. Case $N=1$ is simple enough. To get from $N-1$ to $N$, consider the projection $P(x) := x - \sum_{i<N} x_i F_i(x)$. With this definition, $F_i(P(x)) = 0$ for all $x$. I claim that there is $x$ such that $F_N(P(x)) \neq 0$: otherwise we have $0 = F_N(P(x)) = F_N(x) - \sum_{i<N} F_N(x_i) F_i(x)$, so $F_N$ is linear combination of $F_i$ with $i<N$. So, after rescaling, we can find $x$ such that $F_N(P(x)) = 1$. We can put $x_N = P(x)$, and check it works. This finishes the inductive proof.

Now that we have the sought $x_i$, consider again the projection $P(x) := x - \sum_{i\leq N} x_i F_i(x)$. For any $x$ we have $P(x) \in \ker F_i$, so by the assumption we also have $P(x) \in \ker F$. But now again we have $0 = F(P(x)) = F(x) - \sum_{i<N} F(x_i) F_i(x)$, so $F$ is a combination of $F_i$, as desired.

Edit: The final step is essentially the same as applying the inductively proved claim + supposition that $F$ is not in linear span of $F_i$, to the sequence of functionals $F_1,F_2,\dots,F_N,F_{N+1}:=F$. The $x_{N+1}$ that comes out is in $\bigcap_{i=1}^N \ker F_i$ but not in $\ker F_{N+1} = \ker F$, which is a contradiction.

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    $\begingroup$ Where do you use the orthogonality and the fact of H is a Hilbert space? $\endgroup$
    – FASCH
    May 2, 2013 at 0:34
  • $\begingroup$ I don't. Orthogonality lies in the idea to look for the $x_i$. The proof goes through in the same way for just linear spaces. This "morally" should be the case, because the problem is "finite-dimensional" in its nature (one might restrict to a finitely-dimensional space and work there). $\endgroup$ May 2, 2013 at 6:09

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