(Lieb and Loss Analysis pg. 44)
“A support plane to a graph of a function $f:K\to\mathbb{R}$ at a point $x\in K$ is a plane (in $\mathbb{R}^{n+1}$) that touches the graph at $(x,f(x))$ and that nowhere lies above the graph ... $f$ convex on $K$... at least one support plane at each point of thr interior of $K$. Thus, there exists a vector $V\in\mathbb{R}^n$ (which depends on $x$) such that $$f(y)\ge f(x)+V\cdot (y-x)$$ for all $y\in K$.”
In the line, “and that nowhere lies above the graph” what does he mean? Does he mean that $f=\infty$ at the point where the plane intersects the its graph? Also, how can we conclude $f(y)\ge f(x)+V\cdot (y-x)$ for all $y\in K$?
It means that if $x \in K$ and $(x,y)$ belongs to the support plane then $y \leq f(x)$.
For example, the x-y plane $\{(x,y,0) \mid x, y \in \mathbb R \}$ is a support plane for the graph of the function $f(x,y) = x^2 + y^2$ at the point $(0,0)$.
To answer the follow-up question, I think it will be more clear to slightly change notation. A support plane $W$ to the graph of $f$ at a point $x_0$ must be non-vertical, so there exists a vector $V$ and a scalar $b$ such that $W$ is the graph of the equation $y = V \cdot (x - x_0) + b$. Note that $W$ touches the graph of $f$ at $(x_0,f(x_0))$, which implies that $$f(x_0)= b.$$ Thus, $W$ is the graph of $y = V\cdot (x - x_0) + f(x_0)$. It follows that $$ f(x) \geq V\cdot (x - x_0) + f(x_0)$$ for all $x$ in $K$.