proof of closed graph theorem

Let T:XY be a linear mapping. Denote its graph by G(T), and let p1:X×YX and p2:X×YY be the projections onto X and Y, respectively. We remark that these projections are continuousMathworldPlanetmathPlanetmath, by definition of the productPlanetmathPlanetmathPlanetmath of Banach spacesMathworldPlanetmath.

If T is boundedPlanetmathPlanetmathPlanetmathPlanetmath, then given a sequence {(xi,Txi)} in G(T) which convergesPlanetmathPlanetmath to (x,y)X×Y, we have that




by continuity of the projections. But then, since T is continuous,


Thus (x,y)=(x,Tx)G(T), proving that G(T) is closed.

Now suppose G(T) is closed. We remark that G(T) is a vector subspace of X×Y, and being closed, it is a Banach space. Consider the operator T~:XG(T) defined by T~x=(x,Tx). It is clear that T~ is a bijection, its inversePlanetmathPlanetmathPlanetmath being p1|G(T), the restrictionPlanetmathPlanetmath of p1 to G(T). Since p1 is continuous on X×Y, the restriction is continuous as well; and since it is also surjective, the open mapping theoremMathworldPlanetmath implies that p1|G(T) is an open mapping, so its inverse must be continuous. That is, T~ is continuous, and consequently T=p2T~ is continuous.

Title proof of closed graph theorem
Canonical name ProofOfClosedGraphTheorem
Date of creation 2013-03-22 14:48:47
Last modified on 2013-03-22 14:48:47
Owner Koro (127)
Last modified by Koro (127)
Numerical id 5
Author Koro (127)
Entry type Proof
Classification msc 46A30