Results 1 to 3 of 3

Math Help - Consequence of the closed graph theorem

  1. #1
    Member
    Joined
    Mar 2008
    Posts
    148

    Consequence of the closed graph theorem

    Let U be a closed subspace of L^2(0,1) that is also contained in C([0,1]).

    a) Prove there is an M > 0 such that ||f||_sup <= M||f||_L2 for every f in U.

    b) Prove that for each 0 <= x <= 1 there is a g_x in U such that
    f(x) = <f,g_x> for all f in U and ||g_x||_L2 \leq M.


    It is recommended to use the closed graph theorem. My thinking is that C([0,1]) \subset L^2(0,1) are both Banach spaces, U is a Banach space and the identity map on U from C([0,1]) to L^2(0,1) is closed, linear. Then we get that it is bounded by the closed graph theorem and thus part a) is true.

    b) I am having a hard time convincing myself this is true. For example g_0 would need to be a continuous function with the property that f(0) = <f,g_0> for every f in U.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jun 2009
    Posts
    113
    For b) you could use that a closed subspace of a Hilbert space is again a Hilbert space. For a) you have that the evaluations \delta_x, x\in [0,1] are continuous for the L_2 topology, and obviously linear. Since U with the L_2 norm is Hilbert any element of the dual can be represented in the desired way.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jun 2009
    Posts
    113
    Quote Originally Posted by iknowone View Post

    It is recommended to use the closed graph theorem. My thinking is that C([0,1]) \subset L^2(0,1) are both Banach spaces, U is a Banach space and the identity map on U from C([0,1]) to L^2(0,1) is closed, linear. Then we get that it is bounded by the closed graph theorem and thus part a) is true.
    By the way, the identity map from U \subseteq C([0,1]) to L^2(0,1) is in fact continuous. This implies that the identity U,\|\cdot\|_2)\hookrightarrow (U,\|\cdot\|_\infty)" alt="iU,\|\cdot\|_2)\hookrightarrow (U,\|\cdot\|_\infty)" /> has closed graph, and since (U,\|\cdot\|_2) is closed (and then Banach) this yieds the non trivial inequalty in a), from the first continuity we only obtain \|\cdot\|_2\leq K\|\cdot\|_\infty for certain K positive.

    The key is to observe that T:X \mapsto Y has closed graph whenever T is continuous when Y is endowed with a weaker Hausdorff topology than its natural one (I believe that it is even an equivalence in the locally convex setting). It is a really useful trick in functional Analysis, for checking continuity in a linear mapping from a Banach space to another locally convex space, you only have to check continuity for the weaker known Hausdorff topology in the right hand side. This is the usual application of the closed graph theorem. Above, we only have to consider in U the L^2 topology.

    A more general conclusion from the argument: if X is Banach space with two norms which are comparable, then both are equivalent.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Application of Closed Graph Theorem
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: February 1st 2011, 10:38 AM
  2. Closed Linear Operators/Closed Graph Theorem
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: June 17th 2009, 03:36 AM
  3. Prove continuous -> graph is closed
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 12th 2009, 12:40 AM
  4. Cauchy Mean Value Theorem Consequence problem
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 25th 2008, 02:14 PM
  5. consequence
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 12th 2005, 06:22 PM

Search Tags


/mathhelpforum @mathhelpforum