### Archive

Posts Tagged ‘functional’

## Compact operator maps weakly convergent sequences into strong convergent sequences

Suppose $T \in \mathcal{L}(E,F)$ is a compact operator and $(u_n)$ is a sequence in $E$ such that $u \rightharpoonup u$ (i.e. converges weakly). Prove that $Tu_n \to Tu$ strongly in $F$.

## Slick condition for boundedness of an operator

Consider $E$ a Banach space, and $E^*$ the space of real functionals on $E$. Suppose that $T : E \to E^*$ is a linear operator such that $\langle Tx,x \rangle \geq 0$ for all $x \in E$. Prove that $T$ is bounded.

## Compact & Hausdorff spaces

The following problem wants to prove that compact and Hausdorff spaces have an interesting property. If $(X,\mathcal{T})$ is Hausdorff and compact, and we consider another Hausdorff topology $\mathcal{T}^\prime$ coarser than $\mathcal{T}$ and Hausdorff, then $\mathcal{T}=\mathcal{T}^\prime$.

1) Prove that a compact set in a Hausdorff space is closed.

2) Every continuous function from a compact space onto a Hausdorff space is open.

3) If you have a bijective continuous function from a compact space to a Hausdorff space, then the two spaces are homeomorphic.

A short application, which was my motivation for this post is the following problem:

Let $E$ be a Banach space and $K\subset E$ be a compact subset in the strong topology. Let $(x_n)$ be a sequence in $K$ such that $x_n\to x$ in the weak topology. Prove that $x_n \to x$ strongly.

## Weakly Compact means bounded

Let $E$ be a Banach space, and $A \subset E$ be a compact set in the weak topology. Then $A$ is bounded.

## The complement of a kernel intersection

Let $E$ be a vector space, and $u_1,...,u_n \in E^*$ be linear functionals. Consider $F$ a complement of the space $\bigcap_{1\leq i\leq n}\ker u_i$. Prove that $F$ is finite dimensional and its dimension is at most $n$.

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## Functional is continuous iff kernel is closed

Let $f$ be a linear functional on a topological vector space $X$. Assume $f(x)\neq 0$ for some $x \in X$. Then the following properties are equivalent:

1. $f$ is continuous
2. $\ker f=\{x : f(x)=0\}$ is closed
3. $\ker f$ is not dense in $X$
4. $f$ is bounded in some neighborhood of $0$.

## Linear maximal subspaces & Linear functionals

October 26, 2009 1 comment

Take $X$ to be a vector space over the field $\Gamma$.
1. Let $f:X \to \Gamma$ be a linear functional, which is not identically zero. Then it’s kernel, $\ker f=\{x \in X : f(x)=0\}$, is a linear maximal subspace of $X$.

2. Conversely, given a maximal subspace $X_0$ of $X$, there exists a linear functional $f$ such that $\ker f=X_0$.

3. As a consequence of the above prove that if we have two linear functionals (different from zero identity) $f_1,f_2$ such that $\ker f_1=\ker f_2$ then there exists $\lambda \in \Gamma$ such that $f_1=\lambda f_2$.

As application to the above solve the following:
4. Given $f,g_1,...,g_n$ linear functionals on $X$ such that $\ker f \supset \bigcap\limits_{i=1}^n \ker g_i$ prove that we can find scalars $a_1,...,a_n$ such that $f(x)=a_1g_1(x)+...+a_ng_n(x),\ \forall x \in X$.

Prove that if $f_i,\ i=1..n$ are linearly independent linear functionals on $X$ then there exist elements $x_1,...,x_n \in X$ such that $f_i(x_j)=\delta_{ij}$, where $\delta_{ij}=\begin{cases} 0, & i \neq j \\ 1, & i=j \end{cases}$.