Compact, connected subset of R^n is pathwise connected?

Question

Let $D$ be a compact, connected $Jordan$ domain in $R^n$ with positive volume, and suppose that the fuction $f:D →R$ is continuous. Show that there is a point $x$ in $D$ in which $f(x)=(1/volD)$$\int$$f$ over $D$. (Mean Value Property for integral)

I can prove it when $D$ is pathwise connected

But i have no idea when $D$ is just connected

Is it true compact, connected subset of $R^n$ is pathwise connected?

Or is there any other ways to solve it?

Answer 1

No. The standard example is the Topologist sine curve: $$ \Bigl\{\Bigl(x,\sin\frac{1}{x}\Bigr):0<x\le1\Bigr\}\bigcup\{0\}\times[-1,1]. $$ To prove the MVT all you need is that the continuous image of a connected set is connected.

Answer 2

Explaining better the second question, notice that since $D$ is a compact connected subset of $\mathbb{R}^d$ its image $f(D)$ through continuous function $f$ is also compact and connected. Thus, we have $f(D) = [\min_D f, \max_D f]$ or in other words: $$ \min_D f \le f(x) \le \max_D f, \text{for every $x \in D$.}$$ Just to emphasize, the answer to your first question is yes when $n = 1$ but no when $n \geq 2$. Integrating over $D$ and dividing by $\text{Vol}(D)$ one gets $$ \min_D f \le \frac{1}{\text{Vol}(D)}\int_D f(x) \le \max_D f .$$ But this means that there is $x_0 \in D$ with $f(x_0) = \frac{1}{\text{Vol}(D)}\int_D f(x)$.