necessary and sufficient conditions for a subset to be the graph of a linear operator Let $X$ and $Y$ be two linear vector spaces. Find necessary and sufficient conditions for a subset $G\subset X\times Y$ to be the graph of a linear operator from $X$ into $Y$.
The definition for the graph of a linear operator $T$ with domain $D_T$ is $G=\{(x,Tx):x\in D_T\}$. So I want to look for a different characterization.
I know it is necessary that $G$ is a linear subspace of $X\times Y$. But it is not sufficient, as pointed out by the comment. What is missing?
Thanks!
 A: A somewhat trivial condition is that the intersection of $G$ with the set $\{x\}\times Y$ has to be a singelton for every $x\in X$. This means that
$$D(A):=\{x\in X\colon \exists y\in Y \text{ such that } (x,y)\in G\}$$
and
$$ Ax := y \quad\text{where}\quad (x,y)\in G$$
defines a mapping $A\colon D(A)\subset X\to Y$. Since $G$ is a subspace we always have $$(\lambda x_1+\mu y_1, \lambda x_2+\mu y_2)\in G$$ for $x,y\in G$, i.e. the $A$ defined above is indeed linear.
Note that the above condition is in fact equivalent to the condition that $\{0\}\times Y\cap G = \{0\}$. Since this second condition is obviously weaker than the first one we are left to show that it already implies the first one. Let $(x,y_1), (x,y_2)\in G$. Since $G$ is a subspace it therefore also contains $(x, \frac{1}{2}(y_1+y_2))$ and $(-x, -y_1)$. Hence, again since it is a subspace, it also contains $(0,\frac{1}{2}(y_1-y_2)$. Therefore the condition $\{0\}\times Y=\{0\}$ already implies $y_1=y_2$, i.e. every intersection of $G$ with a set of the form $\{x\}\times Y$ is a singleton.
Similarly, we can see that the condition "$G\cap \{0\}\times Y$ is a singleton" already provides a characterisation, since the subspace $G$ always contains $(0,0)\in X\times Y$.
