Is a closed and bounded convex set with non-empty interior in a normed space homeomorphic to the closed unit ball of this normed space? It is true that for finite dimensional case, a closed and bounded convex set with non-empty interior is just homeomorphic to the unit closed ball. But is the similar conclusion true in an infinitely dimensional normed space? Can anyone give a counter-example if it is not?
 A: Here's my attempt. Let $E$ be a closed and bounded convex set with nonempty interior. WLOG by translation assume that $0\in \operatorname{Int} E$. Define the gauge of $E$ as
$$p(x) = \inf \{t > 0 : x \in tE\}$$
This is well-defined since $E$ contains a ball $B(0,r)$ around the origin so for $x \ne 0$ certainly $x \in \frac{r}{\|x\|}E$. Now it is not hard to show that $p(tx) = tp(x)$ for $t > 0$ and $p(x+y) \le p(x)+p(y).$
Then notice that $\operatorname{Int} E \subseteq B_p(0,1) \subseteq E = \overline{B_p}(0,1)$ where $B_p(0,1)$ is the open unit ball of $p$ and $\overline{B_p}(0,1)$ is the closed unit ball. Indeed:

*

*If $x \in \operatorname{Int} E$, then $1\cdot x = x$ so by continuity of scalar multiplication there is an $\varepsilon>0$ such that $[1-\varepsilon,1+\varepsilon]x \subseteq \operatorname{Int} E$ so $x \in \frac1{1-\varepsilon}$ and hence $p(x) \le \frac1{1-\varepsilon} < 1$ so $x \in B_p(0,1)$.


*If $x \in B_p(0,1)$, then there is a $t \in \langle 0,1\rangle$ such that $x \in tE$. But also $0 \in (1-t)E$ so by convexity we have
$$x = x + 0 \in tE + (1-t)E \subseteq E.$$


*If $x \in E$ then $x \in 1 \cdot E$ so $p(x) \le 1$ and $x \in \overline{B_p}(0,1)$. For the converse inclusion, let $x \in \overline{B_p}(0,1)$. If $p(x) < 1$ then we already know that $x \in E$ so assume $p(x) = 1$. Then for every $\alpha \in \langle 0,1\rangle$ we have $\alpha x \subseteq B_p(0,1) \subseteq E$. For any neighbourhood $V$ of $x$ from $1 \cdot x = x$ by continuity of scalar multiplication there is a $\varepsilon \in \langle 0,1\rangle$ such that $[1-\varepsilon,1+\varepsilon]x \subseteq V$ so $V$ contains points of $E$. Since $V$ was arbitrary it follows that $x \in \overline{E}$.
Now notice that there are constants $m,M > 0$ such that for all $x \in X$ we have
$$m\|x\| \le p(x) \le M\|x\|$$
and in particular $p$ is continuous.

*

*From $B(0,r) \subseteq \operatorname{Int} E \subseteq B_p(0,1)$ we get that $$\|x\| < r \implies p(x) < 1$$
so $p(x) \le \frac2r\|x\|$. From here it follows that $p$ is continuous at $0$ and this implies that $p$ is continuous. Namely, if $x_n \to x$ then $x_n-x\to 0$ and $x-x_n \to 0$ so for any $\varepsilon > 0$ there is a large enough $n\in\Bbb{N}$ so that $p(x_n-x) < \varepsilon$ and $p(x-x_n) < \varepsilon$. The triangle inequality implies
$$p(x) \le p(x-x_n) + p(x_n) \implies p(x)-p(x_n) \le p(x-x_n) < \varepsilon$$
$$p(x_n) \le p(x_n-x) + p(x) \implies p(x_n)-p(x) \le p(x_n-x) < \varepsilon$$
so $|p(x)-p(x_n)| < \varepsilon$ implying $p(x_n) \to p(x)$.

*$E$ is bounded so there is some $R>0$ such that for all $x \in E$ holds $\|x\| \le R$. Hence
$$B_p(0,1) \subseteq E \subseteq B(0,R)$$
which implies that for every $x \in X$ holds $\|x\| < 2R p(x)$.

Now define the bijection $f : E = \overline{B_p}(0,1) \to \overline{B}(0,1)$ as
$$f(x) := \begin{cases} \frac{p(x)}{\|x\|}x, &\text{ if $x \ne 0$}\\
0, &\text{ if $x = 0$}\end{cases}$$
with inverse
$$f^{-1}(y)=\begin{cases} \frac{\|y\|}{p(y)}y, &\text{ if $y \ne 0$}\\
0, &\text{ if $y = 0$}\end{cases}.$$
Continuity of $f$ and $f^{-1}$ at nonzero points follows from continuity of $p$, and for continuity at $0$ we have
$$\|f(x)\| = \frac{p(x)^2}{\|x\|} \le M\|x\| \xrightarrow{x\to 0} 0,$$
$$\|f^{-1}(y)\| = \frac{\|y\|^2}{p(y)} \le \frac1m\|y\| \xrightarrow{y\to 0} 0.$$
Therefore, $f$ is the desired homeomorphism.
