If a function is continuous, then its graph is closed. The converse is false: a counterexample is given by any extension of to the real line.

The Closed Graph Theorem of functional analysis states that a linear map between Banach spaces is continuous whenever its graph is closed. Although the literal extension to nonlinear maps fails, it’s worth noting that linear maps are either continuous or discontinuous everywhere. Hence, if one could show that a nonlinear map with a closed graph has at least one point of continuity, this would be a nonlinear version of the Closed Graph Theorem.

Here is an example of a function with a closed graph and an uncountable set of discontinuities. Let be a closed set with empty interior, and define

For a general function, the set of discontinuities is an set. When the graph is closed, we can say more: the set of discontinuities is closed. Indeed, suppose that a function is bounded in a neighborhood of but is not continuous at . Then there are two sequences and such that both sequences and converge but have different limits. Since at least one of these limits must be different from , the graph of is not closed. Conclusion: a function with a closed graph is continuous at **if and only if** it is bounded in a neighborhood of . In particular, the set of discontinuities is closed.

Furthermore, the set of discontinuities has empty interior. Indeed, suppose that is discontinuous at every point of a nontrivial closed interval . Let ; this is a closed bounded set, hence compact. Its projection onto the -axis is also compact, and this projection is exactly the set . Thus, is closed. The set has empty interior, since otherwise would be continuous at its interior points. Finally, , contradicting the Baire Category theorem.

Summary: for closed-graph functions on , the sets of discontinuity are precisely the closed sets with empty interior. In particular, **every such function has a point of continuity**. The proof works just as well for maps from to any metric space.

However, the above result does not extend to the setting of Banach spaces. Here is an example of a map on a Banach space such that whenever ; this property implies that the graph is closed, despite being discontinuous everywhere.

Let the space of all bounded functions with the supremum norm. Let be an enumeration of all rational numbers. Define the function separately on each subinterval , as

For any two distinct elements of there is a point and a number such that is strictly between and . According to the definition of this implies that the functions and take on different values at the point . Thus the norm of their difference is .

So much for Nonlinear Closed Graph Theorem. However, the space in the above example is nonseparable. Is there an nowhere continuous map between separable Banach spaces such that its graph is closed?