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, that 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 Fσ 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?