Nonlinear Closed Graph Theorem

If a function {f\colon {\mathbb R}\rightarrow {\mathbb R}} is continuous, then its graph {G_f = \{(x,f(x))\colon x\in{\mathbb R}\}} is closed. The converse is false: a counterexample is given by any extension of {y=\tan x} to the real line.

Closed graph, not continuous
Closed graph, not continuous

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 {C\subset {\mathbb R}} be a closed set with empty interior, and define

\displaystyle f(x) = \begin{cases} 0,\quad & x\in C \\ \textrm{dist}\,(x,C)^{-1},\quad & x\notin C \end{cases}

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 {f} is bounded in a neighborhood of {a} but is not continuous at {a}. Then there are two sequences {x_n\rightarrow a} and {y_n\rightarrow a} such that both sequences {f(x_n)} and {f(y_n)} converge but have different limits. Since at least one of these limits must be different from {f(a)}, the graph of {f} is not closed. Conclusion: a function with a closed graph is continuous at {a} if and only if it is bounded in a neighborhood of {a}. In particular, the set of discontinuities is closed.


Furthermore, the set of discontinuities has empty interior. Indeed, suppose that {f} is discontinuous at every point of a nontrivial closed interval {[a,b]}. Let {A_n = G_f \cap ([a,b]\times [-n,n])}; this is a closed bounded set, hence compact. Its projection onto the {x}-axis is also compact, and this projection is exactly the set {B_n=\{x\in [a,b] : |f(x)|\le n\}}. Thus, {B_n} is closed. The set {B_n} has empty interior, since otherwise {f} would be continuous at its interior points. Finally, {\bigcup B_n=[a,b]}, contradicting the Baire Category theorem.

Summary: for closed-graph functions on {\mathbb R}, 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 {\mathbb R^n} to any metric space.


However, the above result does not extend to the setting of Banach spaces. Here is an example of a map {F\colon X\rightarrow X} on a Banach space {X} such that {\|F(x)-F(y)\|=1} whenever {x\ne y}; this property implies that the graph is closed, despite {F} being discontinuous everywhere.

Let {X} the space of all bounded functions {\phi \colon (0,1]\rightarrow\mathbb R} with the supremum norm. Let {(q_n)_{n=1}^\infty} be an enumeration of all rational numbers. Define the function {\psi =F(\phi )} separately on each subinterval {(2^{-n},2^{1-n}]}, {n=1,2,\dots} as

{\displaystyle \psi(t) = \begin{cases} 1 \quad &\text{if } \phi (2^nt-1) > q_n \\ 0 \quad &\text{if } \phi (2^n t-1)\le q_n\end{cases}}

For any two distinct elements {\phi_1,\phi_2} of {X} there is a point {s\in (0,1]} and a number {n\in\mathbb N} such that {q_n} is strictly between {\phi_1(s)} and {\phi_2(s)}. According to the definition of {F} this implies that the functions {F(\phi_1)} and {F(\phi_2)} take on different values at the point {t=2^{-n}(s+1)}. Thus the norm of their difference is {1}.


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

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