# Math Help - Consequence of the closed graph theorem

1. ## 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.

2. 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.

3. Originally Posted by iknowone

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 $iU,\|\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.