Help prove a claim about tangent plane

Question

Given a differentiable function $f(x):\Bbb R^n \to \Bbb R$ and the hyperplane $H=\{ (x,y) \in \Bbb R^{n+1} ~ | \quad a^Tx-y=a^Tx_0-f(x_0)\}$ in $\Bbb R^{n+1}$ which goes through the point $(x_0,f(x_0))$ living on the graph of $f$. Show that if the $\text{epi} f$ is entirely contained in one closed halfspace of $H$, then $H$ must be the tangent plane of $f$ at $x_0$ i.e., $H=\{(x,y) ~|\quad y=f(x_0) + \nabla f(x_0)^T (x-x_0) \}$?

Answer 1

Let assume $$\text{epi} (f) \subseteq H^+ = \{ (x,y)~ | \quad y\ge f(x_0) + a^T (x - x_0) ~ \forall x \in R^n \}$$ Clearly $\text{graph} (f) \subseteq \text{epi} (f) \subseteq H^+ $ thus $$f(x)\ge f(x_0) + a^T (x - x_0) ~~ \forall x \in R^n $$ which is saying that the function $g(x ) = f(x) - f(x_0) - a^T (x - x_0)$ attains its global minimum point at $ x=x_0 $ which implies $\nabla g(x_0) = 0$ so $\nabla f(x_0) = a.$ Hence, $$H = \{ (x,y)~ | \quad y = f(x_0) + \nabla f(x_0) (x - x_0) ~ \forall x \in R^n \}$$