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

Question

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.

Answer 1

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 $i<j$ 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 $\dim(\Sigma)≥k$.