Let $A\subseteq\mathbb{R}^n$ be convex. For any two points $P, Q\in A$, $f(P)<0<f(Q)$. Prove that $\exists$ $ T \in A: f(T)=0$.

Question

A subset $A\subseteq\mathbb{R}^n$ is convex if $\forall$ $P, Q\in A$ the line segment that joins them is contained in $A$.
Let $A \subseteq \mathbb{ R }^n$ be convex, $f: A \to \mathbb{R}$ continuous and for any two points $P, Q\in A$, $f(P)<0<f(Q)$. Prove that $\exists$ $ T \in A: f(T)=0$.

I think I need to use the Intermediate Value Theorem but I am not sure how to start.

Answer 1

Hint: Let $g(t):=f(tP+(1-t)Q)$, with $g:[0,1]\to \mathbb{R}$.

Answer 2

Let's consider the function $g : [0,1] \rightarrow \mathbb{R}$ defined by $$\forall t \in [0,1], \quad g(t)=f(tP+(1-t)Q)$$

This function is continuous and satisfies $f(0)=f(Q) > 0$ and $f(1)=f(P) < 0$. By the intermediate value theorem, there exists $t_0 \in [0,1]$ such that $g(t_0)=0$. So for $T = t_0 P + (1-t_0)Q$, you have $$f(T)=0$$

Answer 3

Give $g: [0,1] \rightarrow \mathbb{R}$ with $g(t) = f(tP+(1-t)Q)$.

Replace $t = 0$ and $t = 1$ in function $g$.

Show that $g (1) <0 <g (0)$.

As $f$ is continuous, $g$ is continuous. So you can apply the Mean Value Theorem (if $g: [a,b] \rightarrow \mathbb{R}$ is continuous and $g(a) < c < g(b)$, then there is a point $t \in (a,b) $ such that $g(t) = c$).

So in this way, you have some point in $(0,1)$ that you use the function $g$ you obtain a a point $T$ on the segment $PQ$ (here we using the convexity) that has a value equal to $f(T) = 0$.