# Tag Info

4

In a metric space you have a way to measure distance and thus the definition of boundedness is, in a sense, obvious. In a topological vector space there are no distances, so of course the definition will be different than that given for metric spaces simply since you can't state the metric definition without a metric. However, if you think about it, the ...

2

A bounded set $S$ in a topological vector space is one for which, given any neighborhood $N$ of the zero vector, there is a scalar $\alpha$ such that $S \subset \alpha N$. For more: http://en.wikipedia.org/wiki/Bounded_set_(topological_vector_space) The motivation for this definition is fairly straightforward--if a "stretching" of a neighborhood of the zero ...

0

One key idea is the connection between balanced absorbing convex sets and (semi-)norms. They are basically one and the same. For every norm the unit ball under that norm is a convex set. Similarly, any balanced bounded absorbing convex set generates a norm via the Minkowski functional (ie: how much do you have to scale the set to hit your point? Define ...

0

To me, the primary relevance of convex spaces is in its very definition: When a space contains two points, it also contains the line segment connecting them. As for convex functions, one justification at calculus level is the uniqueness of local minimum. The most interesting theorem in "real analysis" regarding convex functions is the Jensen's inequality ...

3

Maybe the following can at least partly answer your question. First we look at topological vector spaces in increasing generality: \begin{align*} &\text{finite dimensional vector spaces - Hilbert Spaces - Banach Spaces}\\ &\text{Frechet Spaces - locally convex vector spaces}\\ \end{align*} as Jänich does in his Topology. There he mentions ...

1

Mathematics is always a fine line between imposing conditions that are Strong enough to give you tools with which to show interesting results Weak enough to cover interesting applications Strange enough to not be easily analyzed or trivialized The condition of local convexity happens to be Strong enough to leave you with a rich dual theory, due to the ...

0

Let's take a look at the proof of the Banach space statement you cite. As we will see it easiliy generalizes to Frechet spaces If $T \colon X \to Y$ has a right inverse, say $S \colon Y \to X$, then $TS = \def\Id{\mathord{\rm Id}}\Id_Y$. Let $P := ST$. Then we have $P^2 =(ST)^2 = S(TS)T = ST=P$, so $P$ is a continuous projection. If $x \in \ker P$, then ...

0

So, the continuity of $+\colon X/M \times X/M \to X/M$ is proved in Rudin's functional book and I give a proof of the continuity of the scalar product using the same idea. Then to prove that $.:F \times X/M \to X/M$ is contiunous let $w$ be an open neighborhood of $\vec{0}=M$ in $X/M$. $p^{-1}(w)$ is an open neighborhood of $\vec{0}$ in $X$ since $p$ is ...

1

An other example: the space of the bounded serie provided of the norm of the supremum. We can consider the map $\varphi$ who at the serie $u_0,u_1,u_2,...$ associate the serie $\sin(u_0),u_0,u_1,u_2,...$.

1

I think that $f\colon \mathbb R\to L^1(\mathbb R)$ defined by $$f(x)=\begin{cases}1_{[0,x]}&\text{if}\ x\geq 0\\ -1_{[0,-x]}&\text{if}\ x<0\end{cases}$$ is also a counterexemple.

4

Define $f \colon \mathbb R \to \mathbb R^2$ by $f(x) = (x, \left|x\right|)$. If $\mathbb R$ is endowed with the usual norm given by the absolute value, and $\mathbb R^2$ with the maximum norm $\left|x\right|_\infty = \max\{|x_1|, |x_2|\}$, then $f$ is an isometry.

1

We will show that the second case cannot happen. If there are only finitely many $x_i$ in each $X_k$, we find subsequences $(y_i)$ of $(x_n)$ and $(Y_i)$ of $(X_n)$ (say $y_i = y_{n_i}$, and $Y_i = X_{m_i}$), such that $y_i \in Y_{i+1}\setminus Y_i$. As $Y_i$ is complete, it is closed in $X$, so by Hahn-Banach there is an $x_i^* \in X^*$ such that ...

3

Let $(X,\tau )$ be any Hausdorff topological vector space , and let $\sigma =\{\emptyset , X\}$ be a trivial topology on $X.$ Then the identity $\mbox{id}_X :(X,\tau )\to (X ,\sigma )$ is a continuous and linear bijection but $(X,\sigma )$ is not Hausdorff topological vector space.

Top 50 recent answers are included