# Theorem 4.25 - Rudin's Functional Analysis. Elementary observation about Quotient spaces

I'm studying the theorem in the title and the very first elementary observation is not really clear to me. I think I can state it as follow:

Let $$Y$$ a locally convex space, $$M_0$$ a closed subspace of $$Y$$ and $$\Sigma$$ is the space of all continuous linear functionals on $$Y$$ that annihilate $$M_0$$ then $$\dim Y / M_0 \leq \dim \Sigma$$

I'll quote the book from now on

Pick a positive integer $$k \leq \dim Y / M_0$$. Then there're vectirs $$y_1,\ldots, y_k \in Y$$ such that the vector space $$M_i$$ generated by $$M_0$$ and $$y_1,\ldots y_i$$ contains $$M_{i-1}$$ as proper subspace. By theorem 1.42 each $$M_i$$ is closed and by theorem 3.5 there're functionals $$\Lambda_1, \ldots, \Lambda_k$$ such that $$\Lambda_i y_i = 1$$ but $$\Lambda_i y = 0$$ for all $$y \in M_{i-1}$$. These functionals are linearly indipendent. The following conclusion is therefore reached.

I don't understand and I think the reason is beacause I'm missunderstanding the definition of $$\Sigma$$ which to me is simply $$M_0^{\perp}$$ and whose dimension is at least $$k$$ but it's not really helping me to reach the conclusion, so probably I don't understand the definition of $$\Sigma$$.

Can you clarify why the conclusion is true?

Update : Reference is Theorem 4.25 Rudin's Functional analysis. It's literally the first few lines of the proof of the theorem.

• $\Sigma = \{ f\in Y^* \mid f(y)=0\ \ \forall y\in M_0\}$. If you have that $M_{i-1}$ is closed then the functional $\Lambda_i:M_{i-1} \oplus y_i\cdot \Bbb C\to\Bbb C$, $\Lambda_i(y+\lambda y_i) =\lambda$ is continuous. By Hahn Banach you can extend to a functional on all of $Y$. Commented Nov 19, 2020 at 18:40
• @s.harp Isn't your $\Sigma$ equal to $M_0^{\perp}$? Also I cannot follow your answer. I don't understand how the extension help with the conclusion of the theorem, please write an extended comment or answer. Commented Nov 19, 2020 at 18:48
• Comment of @s.harp explains much better. Don't pay attention to mine. Commented Nov 19, 2020 at 18:48
• @s.harp do you also have a notation mistake? Did you mean $\Lambda_i : M_{i-1} \oplus y_i \to \mathbb{C}$? Despite the extensions anyway I don't really get the inequality between the dimension of the two spaces. Commented Nov 19, 2020 at 18:57

The definition of $$\Sigma$$ is: $$\Sigma= \{ f\in Y^*\mid f(y) =0\ \ \forall y\in M_0\}$$ this agrees with the definition of the annihilator $$M_0^\perp$$ that I know. Now $$M_0$$ is assumed to be closed and $$M_i := M_{i-1}\oplus \Bbb C y_{i}$$ will inductively be closed since you are adding a one-dimensional space to a closed subspace (here we are assuming that $$[y_1],..,[y_k]$$ are linearly independent in $$Y/M_0$$ else you may have $$M_i=M_{i-1}$$).

Now the functional $$\Lambda_i:M_i\to\Bbb C, \qquad (y+\lambda y_i)\mapsto \lambda$$ is well defined.

Further it is continuous: Suppose $$y^{(\alpha)}+\lambda^{(\alpha)}y_i\to y+\lambda y_i$$, by closedness of $$M_{i-1}$$ you get that $$y^{(\alpha)}\to y$$ and $$\lambda^{(\alpha)}\to\lambda$$, whence $$\Lambda_i(y^{(\alpha)}+\lambda^{(\alpha)}y_i)= \lambda^{(\alpha)} \overset{\alpha\to\infty}\longrightarrow \lambda = \Lambda_i(y+\lambda y_i)$$ verifying continuity.

By Hahn-Banach you can extend the $$\Lambda_i$$ to be functionals on all of $$Y$$. Now note that $$\Lambda_i\lvert_{M_0}=0$$ for all $$i$$, hence they are also in $$\Sigma$$. Next see that for $$i you have: $$\Lambda_j(y_i)=0$$ and $$\Lambda_j(y_j)=1$$, you can leverage these two equalities to find that the $$\Lambda_i$$ must all be independent. This gives $$\dim(\Sigma)≥k$$.

• I still don't get it, because by hypothesis we have a $k$ such that $k \leq \dim Y / M_0$ and you essentially proved what I derived myself already, namely $k \leq \dim \Sigma$, how do you combine those to get $$\dim Y / M_0 \leq \dim \Sigma.$$ Unless I'm missing something. Commented Nov 19, 2020 at 19:02
• Well if $\dim Y/M_0$ is finite then let $k=\dim Y/M_0$ and you're done. If its infinite then you have for every $k\in\Bbb N$ that $\dim\Sigma ≥ k$ and $\Sigma$ is also infinite-dimensional. This argument does not prove the inequality if you allow $\dim$ to be "cardinality valued", ie it does not imply if $\dim Y/M_0$ has cardinality $2^{\mathfrak c}$ that then $\dim \Sigma ≥ 2^{\mathfrak c}$, whether that statement is true I don't know. Commented Nov 19, 2020 at 19:06
• I added the reference in my question, this "elementary observation", as the author calls it, it's quite fundamental for the rest of the theorem otherwise to me it's completely nonsense. I'm also noticing in your last comment you have to pick $k = \dim Y / M_0$ while the author seems to pick an arbitrary $k$ less or equal than the dimension, which eventually is infinity. Commented Nov 19, 2020 at 19:09
• The implication is $\dim(\Sigma)≥ k$ for every $k \in \Bbb N$ with $k≤ \dim Y/M_0$. In the event that $Y/M_0$ is infinite you get that $\dim(\Sigma)$ is greater than every finite number, ie also infinite. In the event that $\dim Y/M_0$ is finite take $k=\dim Y/M_0$. Commented Nov 19, 2020 at 19:19
• I missed that bit... that the $k \leq \dim \Sigma$ holds for every $k$ such that $k \leq \dim Y / M_0$ in particular for $k = \dim Y / M_0$ (if finite) otherwise they're both infinite and the inequality holds trivially. Is it right? Commented Nov 19, 2020 at 19:23