# graphs of functions which are closed, but fail to be continuous

I tried looking up a question regarding graphs of continuous functions on this site, but all the ones I found consider functions from $\mathbb{R}$ into $\mathbb{R}$. I have been pondering the following question: given a general topological spaces $X, Y$, and a function $f: X\to Y$, when does $Graph(f)$ closed in $X\times Y$ imply that $f$ is continuous. By the closed graph theorem, this is true whenever $X$ and $Y$ are both Banach spaces.

Also, it is fairly easy to prove that whenever $Y$ is a Hausdorff space and $f$ is continuous, then $Graph(f)$ is closed, but I do not think that the converse is true, so I am trying to find an example where $X$ is some topological space, $Y$ a Hausdorff space, $f: X\to Y$ a function with a closed graph in $X\times Y$, but who fails to be continuous. As of yet I have not been able to find such a counterexample, partially because I have no clue where to look for such a counterexample. I would really appreciate getting some directions to go in.

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Hint: consider the derivative as a function of $C^1[0, 1]$ to $C^0[0, 1]$. This is a linear, unbounded function (hence not continuous), but you can show that its graph is closed. – student Jan 12 '12 at 18:12
@Leandro: what norm are you using on $C^1[0,1]$? If it's $\lVert f'\rVert + \lVert f\rVert$, then $(f_n)$ converges in $C^1[0,1]$ iff both $(f_n)$ and $(f_n')$ converge uniformly on $[0,1]$ – kahen Jan 12 '12 at 18:19
@kahen: the norm on $C^1[0, 1]$ should be just $\|f\|$ (the supremum norm) for this example. – student Jan 12 '12 at 18:21
Your Banach space example needs the hypothesis of "linear" also. – GEdgar Jan 12 '12 at 18:22

$f\colon\mathbb{R}\to\mathbb{R}$ defined by $$f(x)=\frac1x\quad\text{if}\quad x\ne0,\quad f(0)=0.$$
What if we want $X$ to be compact? ie, I need a discontinuous function $f:X \longrightarrow \mathbb{R}$, whose graph is closed. – Janson A.J Nov 17 '15 at 16:55
Ohh.. We just have to take the same function on $[-1,1]$.. – Janson A.J Nov 17 '15 at 17:03
@JansonA.J $X=[0,1]$, same $f$. – Julián Aguirre Nov 17 '15 at 17:05