We have the following definition of a subgradient vector-
We say a vector g ∈ $ R^n $ is a subgradient of f : $ R^n $ → R at x ∈ dom(f) if for all z ∈ dom(f,)
$f(z) ≥ f(x) + g^T (z − x).$
A vector g ∈ $ R^n $ is a subgradient of f at x if and only if (g,−1) defines a supporting hyperplane to epi(f) at (x, f(x)).
My question is that why (g,−1) should define a supporting hyperplane to epi(f) at (x, f(x))? From where did the (g,−1) come from? Is this a point?
A supporting hyperplane for a convex set C in $R^{n}$ is defined by a linear equality of the form
$a^{T}z \leq b$.
If we're given a specific point $\hat{z}$ in $C$ at which the convex set is supported by the hyperplane, then $a^{T}\hat{z}=b$. Thus knowing the point $\hat{z}$ and the vector $a$ is sufficient to find $b$ and determines the supporting hyperplane. We say that the supporting hyperplane at $\hat{z}$ is defined by the vector $a$.
This becomes somewhat more complicated when we consider $\mbox{epi}(f)$, since the epigraph lives in $R^{n+1}$ rather than $R^{n}$.
if $f$ is a function that maps $R^{n}$ to $R$, then the epigraph of $f$ is a set in $R^{n+1}$. If $f$ is convex, and $g$ is a subgradient of $f$ at a specific point $\hat{x}$, then for all $x$
$f(x) \geq f(\hat{x})+g^{T}(x-\hat{x})$.
For any point $(x,t)$ in $\mbox{epi}(f)$,
$t \geq f(x) \geq f(\hat{x})+g^{T}(x-\hat{x})$
Thus a supporting hyperplane for the epigraph of $f$ (in the $n+1$ dimensional $(x,t)$ space!) at the point $(\hat{x}, f(\hat{x}))$ is given by
$g^{T}x-t \leq -f(\hat{x})+g^{T}\hat{x}$
or
$(g,-1)^{T}(x,t) \leq -f(\hat{x})+g^{T}\hat{x}$.
Thus the vector $(g,-1)$ in $R^{n+1}$ together with the constant $-f(\hat{x})+g^{T}\hat{x}$ define a supporting hyperplane for the epigraph of $f$ at the point $(\hat{x}, f(\hat{x}))$.