I know that for a convex function of one variable $f$, it has monotonically increasing difference $\Delta _{t} f( x) :=\frac{f( x+t) -f( x)}{t}$ with its step $t$. And it also have both left and right derivative $f'_{-}( x) \leqslant f'_{+}( x)$ for any inner point in its domain. So I can get that a convex funtion is always bigger than a line go though any point $(x,f(x))$
But what if $f$ is a multivariate convex function?Will it be always bigger than a plane? or more precisely, bigger than a linear( affine) function? I want to proof it naively by the work before. But it seems not a trival inference from function of one variable.
Finally , I want to proof the next propostion:
If $f$ is a multivariate function in $\mathbb{R^n}$(maybe not a smooth function). It also a convex function $$f( \alpha _{1}\boldsymbol{x}_{1} +\dotsc +\alpha _{n}\boldsymbol{x}_{n}) \geqslant \alpha _{1} f(\boldsymbol{x}_{1}) +\dotsc +\alpha _{n} f(\boldsymbol{x}_{n}) ,\alpha _{i} \geqslant 0,\sum\limits _{i=1}^{n} \alpha _{i} =1$$ then there exists a linear transform $\mathscr{A}(\boldsymbol x)=A \cdot \boldsymbol x,A\in \mathbb{R}^{n}$ such that
$$f(\boldsymbol x) >f(\boldsymbol x_0) +A\cdot (\boldsymbol x-\boldsymbol x_{0}) ,\forall \boldsymbol x\in \mathbb{R}^{n}$$