Chapter 17

Let \(E\) be a topological linear space. Prove that a) If Uand Vare open sets, then so is \(U+V=\\{z: z=x+y, x \in U\), \(y \in V\\}\) b) If \(U\) is open, then so is \(\alpha U=\\{z: z=\mathrm{ax}, \mathrm{x} \in U\\}\) provided that \(\mathrm{a} \neq 0\) c) If \(F \subset E\) is closed, then so is \(\alpha F\) for arbitrary a.

Prove that a topological linear space is a \(T_{1}\)-space if and only if the intersection of all neighborhoods of zero contains no nonzero elements.

Prove that a topological linear space \(\mathrm{E}\) automatically has the following separation property: Given any point \(x \in E\) and any neighborhood \(U\) of \(\mathrm{x}\), there is another neighborhood \(\mathrm{V}\) of \(\mathrm{x}\) such that \([\mathrm{V}] \subset U\). Hint. If \(U\) is a neighborhood of zero, then, by the continuity of subtraction, there is a neighborhood \(\mathrm{V}\) of zero such that \(=\) $$ V-V=\\{z: z=x-y, x \in V, y \in V\\} \subset U $$ Suppose \(y \in[V]\). Then every neighborhood of \(y\), in particular \(V+y\), contains a point of \(V\). Hence there is a point \(z \in V\) such that \(z+y \in \mathrm{V}\). It follows that \(\mathrm{y} \in \mathrm{V}-\mathrm{V} \subset U\)

Let \(\mathrm{x}\) be any point of a locally convex topological linear space \(\mathrm{E}\), and let \(U\) be any neighborhood of \(\mathrm{x}\). Prove that \(\mathrm{x}\) has a convex neighborhood contained in \(U\). Hint. It is enough to consider the case \(\mathrm{x}=0\). Suppose \(U\) is a neighborhood of zero. Then there is a neighborhood Vof zero such that \(\mathrm{V}-\mathrm{V} \subset U\), where \(V-V\) is the same as in the hint to Problem 4 . Since \(E\) is locally convex, there is a nonempty convex open set \(V^{\prime} \subset \mathrm{V}\). If \(x_{0} \in \mathrm{V}^{\prime}\), then \(\mathrm{V}^{\prime}-x_{0}\) is a convex neighborhood of zero contained in \(U\).

Given a linear space \(\mathrm{E}, \mathrm{a}\) set \(U \subset \mathrm{E}\) is said to be symmetric if \(\mathrm{x} \in U\) implies \(-x \in U .\) Let \(\mathscr{B}\) be the set of all convex symmetric subsets of \(E\) such that each coincides with its own interior. Prove that a) \(\mathscr{B}\) is a system of neighborhoods of zero determining a locally convex topology \(\tau\) in \(\mathrm{E}\) which satisfies the first axiom of separation; b) The topology \(\tau\) is the strongest locally convex topology compatible with the linear operations in \(E\); c) Every linear functional on \(\mathrm{E}\) is continuous with respect to \(\tau\).

Two norms \(\|\cdot\|_{1}\) and \(\|\cdot\|\), in a linear space \(\mathrm{E}\) are said to be compatible if, whenever a sequence \(\\{x\), in \(E\) is fundamental with respect to both norms and converges to a limit \(\mathrm{x} \in \mathrm{E}\) with respect to one of them, it also converges to the same limit \(\mathrm{x}\) with respect to the other norm. A linear space E equipped with a countable system of compatible norms \(\|\cdot\|_{n}\) is said to be countably normed. Prove that every countably normed linear space becomes a topological linear space when equipped with the topology generated by the neighborhood base consisting of all sets of the form $$ U_{r, \varepsilon}=\left\\{x: x \in E,\|x\|_{1}<\varepsilon, \ldots,\|x\|_{r}<\varepsilon\right\\} $$ for some number \(\varepsilon>0\) and positive integer \(\mathrm{r}\).

Problem 14. Prove that every countably normed space satisfies the first axiom of countability. Hint. Replace the system of neighborhoods \(U_{r, \varepsilon}\) by the subsystem such that \(E\) takes only the values $$ 1, \frac{1}{2}, \ldots, \frac{1}{n}, \ldots $$ (this can be done without changing the topology). Comment. Thus the topology in E can be described in terms of convergent sequences (recall Sec. 9.4).

Prove that the topology in a countably normed space can be specified in terms of the metric $$ \rho(x, y)=\sum_{n=1}^{\infty} \frac{1}{2^{n}} \frac{\|x-y\|_{n}}{1+\|x-y\|_{n}} \quad(x, y \in E) $$ First verify that \(\rho(x, y)\) has all the properties of a metric, and is invariant under shifts in the sense that \(\rho(x+z, y+z)=\rho(x, y)\) for all \(x, y, z \in E\). Comment. A countably normed space is said to be complete if it is complete with respect to the metric (3).

The norms \(\|\cdot\|_{n}\) in a countably normed space \(E\) can be assumed to satisfy the condition $$ \|x\|_{k} \leqslant\|x\|_{l} \quad \text { if } \quad k