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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.

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  • $\begingroup$ 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 ? $\endgroup$
    – nicomezi
    Feb 3, 2020 at 15:11
  • $\begingroup$ Looks fine to me. $g$ is a bijection between $[0,1]$ and $[a,b]$. $\endgroup$
    – copper.hat
    Feb 3, 2020 at 15:11
  • $\begingroup$ @nicomezi: Why not? $\endgroup$
    – copper.hat
    Feb 3, 2020 at 15:15
  • $\begingroup$ Yes, you are correct, I will fix it when I get to my laptop $\endgroup$
    – copper.hat
    Feb 3, 2020 at 15:18
  • $\begingroup$ As @nicomezi pointed out, the bijection is $t \mapsto ta + (1-t)b$. $\endgroup$
    – copper.hat
    Feb 3, 2020 at 16:05

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$ Feb 3, 2020 at 17:28
  • $\begingroup$ 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 $g(c) = y$. $\endgroup$ Feb 3, 2020 at 17:35
  • $\begingroup$ 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 $f(x) = y$. $\endgroup$ Feb 3, 2020 at 17:37

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