$X$ is a normed vector space and $T:X\to X$ is a function that has a closed graph, does $T$ map closed sets to closed sets? Here is the question:
Suppose that $X$ is a normed vector space and $T:X\to X$ is a function that has a closed graph. Is it true that $T$ maps closed sets to closed sets? Is it true if $T$ is linear?

So I think I am over thinking this one, but I am having some issues getting started with this. I am a bit unsure if I should be working directly with the definition of a graph that I have been given:

When $X$ and $Y$ are normed linear spaces and $T:X\to Y$ is a linear map, the graph of $T$ is $\{(x,Tx):x\in X\}$. It should be noted that the graph of $T$ is a subset of $X\times Y$.

Or if this can be worked out from the Closed Graph Theorem:

If $X$ and $Y$ are Banach spaces and $T:X\to Y$ is linear, then $T$ is bounded if and only if $graph(T)$ is closed in $X\times Y$.

I am thrown off though, as I do not have Banach spaces to work with, and in the first case $T$ is not linear, and therefore the assumptions of the Closed Graph Theorem are not met...
Any guidance as to where to begin, would be appreciated.
 A: Consider $T:\mathbb R^2 \to \mathbb R^2$ defined by $T(x,y) = (x + \sqrt{2} y,0)$, and the closed set $\mathbb Z^2$.
A: Let $X=C[0,1]$ and define $T : C[0,1]\rightarrow C[0,1]$ by $Tf = \int_{0}^{x}f(t)\,dt$. $X$ is a Banach space, and $T$ has a closed graph because $T$ is a bounded linear operator. And $C[0,1]$ is closed, but $T C[0,1]$ is not closed in $C[0,1]$.
To see that $TC[0,1]$ is not closed, first notice that $T$ is injective because $Tf=0$ implies $f=0$ by the Fundamental Theorem of Calculus. So, if the range of $T$ were closed, then it would have a bounded inverse by the Closed Graph Theorem, which would give the existence of a constant $m > 0$ such that $\|Tf\| \ge m\|f\|$ for all $f \in C[0,1]$. To see that such a constant $m > 0$ cannot exist, define
$$
               f_{n}(t) = \left\{\begin{array}{cc}
                                    1-nt, & 0 \le t \le 1/n \\
                                     0, & 1/n \le t \le 1
                                 \end{array} \right..
$$
Then $\|f_{n}\|=1$ for all $n=1,2,3,\cdots$, and $\|Tf_{n}\|=1/(2n)$.
