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Let $f:\mathbb{R}^n \rightarrow \mathbb{R} $ be a continuous function and let $a,b \in \mathbb{R}^n$. Let $g:\mathbb{R} \rightarrow \mathbb{R} $ be defined as:

$ g(t) = f(ta + (1-t)b) $.


i) Prove that $g$ is continuous.

My answer: I have done this using the fact that g is a composition of continuous functions and the fact that f is continuous

ii) Prove that the following Intermediate Value Theorem is true for $f$:

If $d$ is a number between $f(a)$ and $f(b)$ then there exists a point $c \in \mathbb{R}^n $ such that $f(c)=d$.

My attempt at a proof:

First we make the observation that $g(0)=f(b)$ and that $g(1)=f(a)$.

Since $g$ is a real and continuous function $\implies$ there exists a $c' \in[0,1]$ such that $g(c')=d$ since d is a number between $f(a)$ and $f(b)$. Thus $f(c'a+(1-c')b)=d$ and were are done.

iii) State and prove a similar Intermediate Value Theorem for continuous functions $h:R \rightarrow \mathbb{R}$ and $h':B \rightarrow \mathbb{R}$ where $R$ is an open or closes rectangle in $\mathbb{R}^2$ and B is an open or closed ball in $\mathbb{R}^3$

My thoughts: Since we have proven it for $\mathbb{R}^n$ is the arguement not just to say that both sets are subsets of $\mathbb{R}^n$?


I would like to know if the answer to iii) is that simple and also if my proof in ii) is correct? Thanks in advance!

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I would like to know if the answer to iii) is that simple and also if my proof in ii) is correct? Thanks in advance!

No, you have to do a little bit more, because for given $a,b\in R$ and $d$ between $h(a)$ and $h(b)$, you have to check that $c$ is still in $R$ and not in $\mathbb R^2\setminus R$. But using the same ansatz as above, you have just to do a slight change to get a $c\in R$ such that $h(c)=d$.

Your proof i) and ii) is correct, but at ii) you should define $c:=c'a+(1-c')b$ at the end to get the $c$.

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