Archive

Posts Tagged ‘functional’

Compact operator maps weakly convergent sequences into strong convergent sequences

October 6, 2011 Leave a comment

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.

Read more…

Slick condition for boundedness of an operator

September 27, 2011 Leave a comment

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.

Read more…

Compact & Hausdorff spaces

March 1, 2011 Leave a comment

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.

Read more…

Weakly Compact means bounded

February 28, 2011 1 comment

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

Read more…

The complement of a kernel intersection

February 18, 2011 Leave a comment

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.

Categories: Uncategorized Tags:

Functional is continuous iff kernel is closed

January 27, 2011 2 comments

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.

Read more…

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}.