Explaining maths

The Ratio Test is a powerful test for both sequences and series of non-zero real numbers. It comes in various forms, but here is one commonly used version. Theorem (Ratio Test) Let $(x_n)_{n\in\N}$ be a sequence of non-zero real numbers. Suppose that \[\frac{|x_{n+1}|}{|x_n|}\to L \text{ as } n\to\infty\,,\] where $L$ is either a non-negative real number or $+\infty$. (a) If $L>1$, then $|x_n| \to +\infty$ as $n \to + \infty$, the sequence $(x_n)$ is divergent, and the series $\displaystyle \sum_{n=1}^\infty x_n$ diverges. (b) If $L \in [0,1)$, then the series $\displaystyle \sum_{n=1}^\infty x_n$ is absolutely convergent (and hence convergent), and the sequence $(x_n)$ converges to $0$. (c) If $L=1$, then the Ratio Test is inconclusive , and you need to use a different test . (The Ratio Test tells you nothing in this case about the convergence or otherwise of either the sequence or the series.) Notes Here (b) does not tell you anything about the value of the sum of the series ...

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 ...

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 $E_R$. H...

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 conti...

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 $R:E^*\to F^*$ defin...