If a map between separable Banach spaces has closed graph, does it have a point of continuity? 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
$$
y(t) = \begin{cases} 1 \quad &amp;\text{if   $f(2^nt-1)&gt;q_n$} \\ 
0 \quad &amp;\text{if  $f(2^n t-1)\le q_n$}
 \end{cases} 
$$
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.
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 .$
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 .$