Theorem 5.9 Rudin's Real and Complex analysis

Question

I'm trying to understand the proof the of the following theorem:

The Open Mapping Theorem Let $U, V$ be the unit open balls of the Banach spaces $X$ and $Y$. To every bounded linear transformation $\Lambda$ of $X$ onto $Y$ there's a $\delta > 0$ such that $$ \delta V \subset \Lambda (U) $$

I report the proof, with my questions.

Proof : Given $y \in Y$ there's an $x \in X$ such that $\Lambda x = y$;

This clearly follows from the onto of the statement of the theorem, right?

if $\left\lVert x \right\rVert < k$, it follows that $y \in \Lambda(kU)$. Hence $Y$ is the union of the sets $\Lambda(kU)$, for $k = 1,2,3 ...$. Since $Y$ is complete, the Baire Theorem implies that there's a non empty open set $W$ in the closure of some $\Lambda(kU)$.

I believe Baire theorem is applied as follows: Since a complete metric space is not of first category it means it cannot be a countable union of nowhere dense, implying that there's $k$ such that $\Lambda(kU)$ is not a nowhere dense, which implies that for such $k$ there's a non empty openset $W \subseteq \bar{\Lambda}(kU)$. Is it correct?

This means that every point of $W$ is the limit of a sequence $\left\{ \Lambda x_i \right\}$, $x_i \in kU$. From now on, $k$ and $W$ are fixed. Chose $y_0 \in W$, and choose $\eta > 0$, so that $y + y_0 \in W$ if $\left\lVert y \right\rVert < \eta$. For such $y$ there're sequences $\left\{ x_i' \right\}, \left\{ x_i'' \right\}$ such that $$ \Lambda x_i' \to y_0, \;\;\; \Lambda x_i'' = y_0 + y \;\;\;, i \to \infty $$ Setting $x_i = x_i'' - x_i'$, we have $\left\lVert x_i \right\rVert < 2k$ and $\Lambda x_i \to y$. Since this hold for every $y$ with $\left\lVert y \right\rVert < \eta$, the linearity of $\Lambda$ shows that the following is true if $\delta = \eta/2k$

To each $y \in Y$ and to each $\epsilon > 0$ there's an $x \in X$ such that $$ \left\lVert x \right\rVert < \delta^{-1}\left\lVert y \right\rVert \;\;\; \text{and} \;\;\; \left\lVert y - \Lambda x \right\rVert < \epsilon \;\;\; (3) $$

I can't really figure how this bit follows from the linearity and our assumption on $\delta$, can you clarify?

I can't also quite figure what's the purpose of the final part of the proof

Fix $y \in \delta V$, and fix $\epsilon > 0$. By (3) there's an $x_1$ with $\left\lVert x_1 \right\rVert < 1$ and $$ \left\lVert y - \Lambda x \right\rVert < \frac{1}{2}\delta \epsilon $$

Why do we have $\left\lVert x_1 \right\rVert < 1$?

Suppose $x_1,\ldots,x_n$ are chosen so that $$ \left\lVert y - \Lambda x_1 - \ldots - \Lambda x_n \right\rVert < 2^{-n}\delta \epsilon \;\;\; (5) $$ Use (3), with $y$ replaced by the vector on the left side of (5), to obtain an $x_{n+1}$ so that (5) holds with $n+1$ in place of $n$, and $$ \left\lVert x_{n+1} \right\rVert < 2^{-n}\epsilon $$

I think all the rest will be clear once someone sheds some lights on my questions.

Answer 1

I don't see any problems with your application of surjectivity and also with the application of Baire category theorem the way you described it.

I can't really figure how this bit follows from the linearity and our assumption on $\delta$, can you clarify?$\newcommand{\norm}[1]{\|#1\|}$

Maybe I am missing something, but I am only able to show this if we assume $\norm y\le\eta$. (Still, it is true that there exists $\eta>0$ such that closed $\eta$-ball around $y_0$ is inside $W$.) EDIT: However, see Daniel Fischer's comment for a way to do this with strict inequality.

I.e., I will try to work as if this part of the proof says: "choose $\eta > 0$, so that $y + y_0 \in W$ if $\norm{y}\le\eta$".

Now, suppose we are given arbitrary $y\in Y$. Let us denote $$\widetilde y=\frac{\eta}{\norm{y}}y.$$ Since $\norm{\widetilde y}=\eta$, we have that there exists a sequence of points $x_i$ such that $\norm{x_i}<2k$ and $\Lambda x_i\to \widetilde y$. In particular, we can choose $\widetilde x$ such that $\norm{\widetilde y-\Lambda\widetilde x}<\frac\eta{\norm y} \varepsilon$.

Now if we denote $C=\frac{\norm y}\eta$ and take $$x=C \widetilde x$$ then we have $$\norm{y-\Lambda x} = \norm{C(\widetilde y-\Lambda\widetilde x)}= C\norm{\widetilde y-\Lambda\widetilde x}<\varepsilon.$$ At the same time $$\norm x=C\norm{\widetilde x}<\frac1\delta\cdot C\norm{\widetilde y}=\frac1\delta\norm{y}.$$

Why do we have $\left\lVert x_1 \right\rVert < 1$?

If we are using $(3)$ for $y\in\delta V$, that means that $\norm y<\delta$ then we have $$\norm{x}\le \frac1\delta\norm{y} < 1.$$