I've been asked to construct a proof for the intermediate value theorem in higher dimensions, $\mathbb{R}^n$.
Setup
Let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ be a continuous function. Let a and b be points in $\mathbb{R}^n$.
Define the "help" function $g(t)=f(ta+(1-t)b)$
Assigment
Prove that if d is a number such that $f(a)<d<f(b)$, then there exists a point $c\in\mathbb{R}^n$ such that $f(c)=d$.
Give specific proofs in the cases where f is defined on a rectangle in $\mathbb{R}^2$ and a ball in $\mathbb{R}^3$
Solution (Attempted)
g is a (continuous) function in $\mathbb{R}$, which means we can use the "normal" Intermediate value theorem (proof already given in our book). So for any d between $g(t)$ and $g(t')$ there is a c such that $g(c)=d$.
We note that $g(0)=f(b)$ and $g(1)=f(a)$. So $g[0,1]=[f(b),f(a)]$.
Is this sufficient? I feel that it's enough to conclude that f attains any value d between $f(a)$ and $f(b)$.
Sadly, I am rather clueless on the issue, when f's domain is narrowed to being a rectangle, or a ball.
Best Answer
Your proof is okay apart from some writing stuff $($like $[f(b),f(a)]$ when we may have $f(a) < f(b))$.
It does not matter that $f$'s domain is a rectangle or a ball, provided they're filled. What matters for this proof is that the segment $ta+(1-t)b$ lie in the domain.
More generally, what matters here is continuity and connectedness. The intermediate value theorem is a consequence of two general facts: continuous functions map connected sets to connected sets, and connected sets of the real line are intervals.