Explaining maths

In this post we sketch a second proof of the Hahn-Banach Extension Theorem for continuous linear functionals on real normed spaces, this time using transfinite induction. We omit some of the details, which the reader can fill in. In particular, some of the details are similar to those involved in proving the claim made in Part V (when we obtained an upper bound for a chain). For a non-zero ordinal $\gamma$, recall that, with ordinal interval notation, we have $\gamma=[0,\gamma)$. That is, $\gamma$ is equal to the set of all ordinals which are strictly less than $\gamma$. However, we use ordinal interval notation throughout. If you prefer to use $\gamma$ instead of $[0,\gamma)$ that is fine, of course! Looking at the Wikipedia page on the Hahn-Banach Theorem , apparently both Hahn and Banach (independently) used transfinite induction in their proofs, rather than Zorn's Lemma. Theorem (Hahn-Banach Extension Theorem for linear functionals on normed spaces over $\R$)  Let $(E,\jnorm)$

In this post we prove the general Hahn-Banach Extension Theorem for continuous linear functionals on real normed spaces. (As we saw in Part IV, the result for complex normed spaces then follows from this.) Here we give the usual proof using Zorn's Lemma. In the next post we will give a second proof using transfinite induction, similar to the proof we gave for separable normed spaces. Theorem (Hahn-Banach Extension Theorem for linear functionals on normed spaces over $\R$)  Let $(E,\jnorm)$ be a normed space over $\R,$ and let $F$ be a linear subspace of $E$. Let $\psi \in F^*$. Then there exists $\phi \in E^*$ with $\|\phi\|_\op=\|\psi\|_\op$ and such that $\phi$ is an extension of $\psi$. Proof using Zorn's Lemma Let set $S$ to be the set of all ordered pairs $(Y,\theta)$ of the following form: $Y$ is a linear subspace of $E$ with $F \subseteq Y$ and $\theta \in Y^*$, $\theta|_F=\psi$, and $\|\theta\|_\op = \|\psi\|_\op.$ Certainly  $(F,\psi) \in S$, and so $S \neq \emptyset$.

In this post we look at how every complex normed space can also be regarded as a real normed space, and how this fits with bounded linear functionals and the Hahn-Banach theorem. Let $(E,\jnorm)$ be a normed space over $\C$. Then $E$ is also a normed space over $\R$, with the same addition, but only considering multiplication by real scalars. This has no effect on the metric induced by $\jnorm$ or the topology. In particular, separability is not affected by this. Still, let us denote the real version by $E_\R$, so that $(E_\R,\jnorm)$ is the corresponding real normed space. If $E$ is finite-dimensional (over $\C$), then $E_\R$  is finite-dimensional (over $\R$), but the dimension doubles. Indeed, if $E$ is $n$-dimensional over $\C$ with basis $x_1,x_2,\dots,x_n$ (over $\C$), then it is easy to see that $x_1, \i x_1, x_2, \i x_2, \dots, x_n, \i x_n$ is a basis for $E_\R$ (over $\R$). Now let $F$ be a (complex) linear subspace of $E$. Then $F_\R$ is a real linear subspace of $E_R$. Howev

In this post we will see how the Hahn-Banach key lemma  from Part II allows us to prove the Hahn-Banach theorem for extensions of bounded linear functionals on separable normed spaces, without the need to use Zorn's Lemma (or equivalent). First, we recall an easy result about extensions of bounded linear functionals to the closure of a linear subspace. Let $\F$ be $\R$ or $\C$, and let $(E,\jnorm)$ be a normed space over $\F.$ Now let $F$ be a linear subspace of $E$. Then the closure of $F$ in $E$, $\overline{F}$, is also a linear subspace of $E$. As usual, we give $F$ and $\overline{F}$ the norm obtained by restricting $E$'s norm, $\jnorm$. Proposition 1 (extensions of linear functionals to the closure of a subspace) Let $\psi \in F^*$. Then $\psi$ has a unique extension $\tilde \psi \in {\overline{F}\,}^*$ (the dual space of $\overline{F}$), and we have \[\|\tilde\psi\|_\op = \|\psi\|_\op\,.\] Comments This is a special case of a more general result about extensions of contin

In this post we establish (in Lemma 2 below) the "key step" needed for the Hahn-Banach theorem (as described in the previous post), where we extend our linear functional by one real dimension. First we note an elementary fact about the operator norm for continuous linear functionals on real normed spaces. Lemma 1 (Operator norm for real linear functionals) Let $(E,\jnorm)$ be a normed space over $\R$, and let $\phi \in E^*.$ Then $(1)\Jdisplay\|\phi\|_\op = \sup \{\phi(x): x \in E, \|x\| \leq 1\,\}\,.$ Comment This looks very like the definition of the operator norm! However the subtle difference is that we work with $\phi(x)$ instead of $|\phi(x)|$ here. Our actual definition of the operator norm is that \[\|\phi\|_\op = \sup \{|\phi(x)|: x \in E, \|x\| \leq 1\,\}\,.\] Note that this would make no sense, as it stands, for complex normed spaces (though we could work with the real part of $\phi(x)$ in that case, and we would again obtain a true result). Let us refer to the

In this series of posts (intended for mathematics undergraduate students in their 3rd or 4th year) I am going to focus on the Hahn-Banach extension theorem for bounded linear functionals on normed spaces. I may also say something about more general versions in terms of sublinear functions.  Let $\F$ be $\R$ or $\C$, and let $(E,\jnorm)$ be a normed space over $\F$. We denote by $E^*$ the set of continuous (bounded) linear functionals on $E$, that is, the set of continuous linear maps from $(E,\jnorm)$ to $(\F,|\cdot|).$ Since $(\F,|\cdot|)$ is complete, it follows that $E^*$ is always complete when given the operator norm  $\jnorm_\op$ defined, for $\phi\in E^*$, by \[\|\phi\|_\op = \sup\{|\phi(x)|:x \in E, \|x\|\leq 1\,\}\,.\] Now let $F$ be a linear subspace of $E,$ and give $F$ the norm obtained by restricting $\jnorm$ to $F.$ We still denote this restricted norm by  $\jnorm$. We now have another dual space $F^*,$ along with a restriction map $R:E^*\to F^*$ defined by $R(\phi)=\phi

I think this will be the last post in this series. This time we will look at the special case of power series, using Theorem 5 from the previous part to help us to justify differentiation term by term. Notation:  Recall that we denote the set  \(\N \cup \{0\}\) of  non-negative  integers by \(\N_0.\) Recall the statement of our Theorem 5. Theorem 5 (Term by term differentiation of series) Let $I$ be a nondegenerate interval in $\R$, and let $M_n$ ($n \in \N_0$) be non-negative real numbers such that $\sum_{n=0}^\infty M_n < \infty\,.$ Let $f_n$ ($n \in \N_0$) be differentiable functions from $I$ to $\R,$ and suppose that, for all $n \in \N_0$ and all $x \in I,$ we have $|f_n'(x)|\leq M_n.$ Suppose further that, for  all $x \in I$, the series $\sum_{n=0}^\infty f_n(x)$ converges. Define $f:I \to \R$ by \[ f(x)=\sum_{n=0}^\infty f_n(x)\,.\] Then $f$ is differentiable on $I$, and, for each $x \in I$, \[f'(x)=\sum_{n=0}^\infty f_n'(x)\,.\] We are now ready to prove our main