It is well known that the closed graph theorem does not directly extend to nonlinear maps: even for functions from $\mathbb{R}$ to $\mathbb{R}$, having closed graph does not imply continuity. But let's consider the following reformulation of the closed graph theorem:

If a linear map between Banach spaces has closed graph, it has a point of continuity.

This does not essentially change the meaning, since a linear map is either everywhere continuous or everywhere discontinuous. But now there is a better chance of nonlinear generalization. Hence, my question:

Suppose that a nonlinear map between separable Banach spaces has closed graph. Does it necessarily have a point of continuity?

Nonseparable counterexample

Since the above seems too good to be true, I tried to find a counterexample. So far, found it only in nonseparable setting.

Let $X$ the space of all bounded functions $x:(0,1]\to\mathbb R$ with the supremum norm. Let $(q_n)_{n=1}^\infty$ be an enumeration of the rationals. Define the function $y=F(x)$ separately on each subinterval $(2^{-n},2^{1-n}]$, $n=1,2,\dots$ as
$$ y(t) = \begin{cases} 1 \quad &\text{if $f(2^nt-1)>q_n$} \\ 0 \quad &\text{if $f(2^n t-1)\le q_n$} \end{cases} $$ I claim that the map $F:X\to X$ satisfies $\|F(x_1)-F(x_2)\|=1$ whenever $x_1\ne x_2$. Hence, it is nowhere continuous and its graph is a discrete (in particular closed) set.

Indeed, since $x_1\ne x_2$, there is a point $s\in (0,1]$ and a number $n\in\mathbb N$ such that $q_n$ is strictly between $x_1(s)$ and $x_2(s)$; according to the definition of $F$ this implies that the functions $F(x_1)$ and $F(x_2)$ take on different values at the point $t=2^{-n}(s+1)$.

Finite-dimensional case

In finite dimensions, a map with a closed graph is continuous outside of a closed set with empty interior. The proof is not very interesting, so I link to it instead of adding it to the post.

  • $\begingroup$ For finite dimensional, in some compact ball there must be infinitely many points of the graph, and an accumulation point of it would give a continuity point. $\endgroup$
    – Carol
    Feb 9 '15 at 2:55
  • $\begingroup$ Just an observation. But what I am thinking is to take a closed ball (now in the infinite dimensional separable case) that contains uncountably many points of the graph (there must be one) and start slicing it with increasing finite-dimensional subspaces. These slices cannot have all finitely many points of the graph. Then use one finite-dimensional slice ofthe ball that contains infinitelly many points of the graph to get the continuity point in the same manner. $\endgroup$
    – Carol
    Feb 9 '15 at 2:58
  • 2
    $\begingroup$ By the way all infinite dimensional separable Banach spaces are homeomorphic, so you can reduce your problem to the case of non-linear map on $\ell_2$ $\endgroup$
    – Norbert
    Jul 25 '16 at 20:44
  • 2
    $\begingroup$ Interesting that the Bounty setter had their account deleted, so I guess only a 50% autoaward is possible. $\endgroup$ Jul 30 '16 at 18:30

A variation on the non-separable counter-example.
Let $X$ as above, be the space of all bounded functions $x:(0,1]→\mathbb(R$ with the supremum norm. Since $\#(\ell_2(\mathbb N)) = \#(0,1]$, there is a bijection $g: \ell_2(\mathbb N) \rightarrow (0,1]$. Define $F: \ell_2(\mathbb N) \rightarrow X$ as $ F(\phi) = I_{(0,g(\phi)]} ,$ the indicator function of the interval $(0,g(\phi)]. $ As abobe, $\|F(\phi_1)-F(\phi_2)\|=1 $ if $ \phi_1 \ne \phi_2 .$


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