# Intermediate Value Theorem for higher dimensions

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 $$f(a), then there exists a point $$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.

• What you wrote is enough to me. I am not sure I understand what is being asked by "specific proofs" for the other cases, but can you imagine domains where it does not work ? Feb 3, 2020 at 15:11
• Looks fine to me. $g$ is a bijection between $[0,1]$ and $[a,b]$. Feb 3, 2020 at 15:11
• @nicomezi: Why not? Feb 3, 2020 at 15:15
• Yes, you are correct, I will fix it when I get to my laptop Feb 3, 2020 at 15:18
• As @nicomezi pointed out, the bijection is $t \mapsto ta + (1-t)b$. Feb 3, 2020 at 16:05

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.
• Well yeah, it did seem very backwards to write $g[0,1]=[f(b),f(a)]$. But I'm not entirely sure how to fix it. Notation wise. Do you just say "assume $f(a)<f(b)$ and then $g[0,1]=[f(a),f(b)$ without loss of generality? Feb 3, 2020 at 17:28
• Notice that while $\{g(0),g(1)\} = \{f(a),f(b)\}$, for $x\in(0,1)$ one might even have $g(x)$ outside of $[f(a),f(b)]$ or $[f(b),f(a)]$. Regardless of whether $g(0) = f(a)$ or $g(0) = f(b)$, regardless of whether $f(a)<f(b)$ or $f(a) > f(b)$, and regardless of whether $g([0,1]) = [f(a),f(b)]$ or strictly contains it, the theorem holds. The usual way to say it with words: for each $y$ between $f(a)$ and $f(b)$, there is some $c$ between $0$ and $1$ with $g(c) = y$. Feb 3, 2020 at 17:35
• For instance, consider $f(x) = x^2$. With $a = -1$ and $b = 2$ we have $f(a) = 1$ and $f(b) = 4$. But we don't have $f([a,b]) = [1,4]$, we have $f([a,b]) = [0,4]$. The theorem still holds, and for each $y\in(1,4)$ we can find $x\in(-1,2)$ with $f(x) = y$. Feb 3, 2020 at 17:37