Extending the closed unit ball in an open set

Question

In a normed space (over the reals), let $C_1 = \{x : \|x\| \leq 1\}$ the closed unit ball, and let $U$ be an open set including $C_1$.

In $\mathbf{R}^p$, it is intuitively apparent that you can "extend" the closed unit ball so that $C_{1+\epsilon} = \{x : \|x\| \leq 1 + \epsilon \} \subset U$. I managed to prove this for all finite-dimensional spaces.

Here is the proof: We know $C = \{x : \|x\| = 1 \}$ is compact. For each point $x \in C$, take two open balls of radius $r_x$ and $r_x/2$ in $U$ containing $x$. By compactness we can have a finite collection of open balls of radius $r_{x_n}/2$ covering $C$, and included in $U$. Take the minimum of $r_{x_n}$ as $r$.

For any $x$ with $1 \leq \|x\| \leq 1 + r/2$, we have $x/\|x\| \in C$, and thus $x/\|x\|$ is in some $V_n$. Computation gives us $\|x - x/\|x\|\| = \|x\|(1 - 1/\|x\|) < r/2$, and thus $\|x - x_n\| \leq \|x - x/\|x\|\| + \|x/\|x\| - x_n\| < r/2 + r/2 = r$, and thus $x \in U$.

The above proof made essential usage of the compactness of the boundary of the closed unit ball, which I know is to be false in infinite-dimensional spaces. I would like to know if a similar result holds in infinite-dimensional normed spaces, or if there exists a counterexample (which I believe is the case, but have been unable to find one).

Answer 1

Consider the vector space $E=(\mathcal{C}^0_{\rightarrow 0}(\mathbb{R}, \mathbb{R}), ||.||_{\infty})$ of real-values continous functions on $\mathbb{R}$ that tends to $0$ in $\pm \infty$, endowed with the sup-norm.

Then the closed unit ball is $B=\lbrace f \in \mathcal{C}^0_{\rightarrow 0}(\mathbb{R}, \mathbb{R}) \mid \forall x \in \mathbb{R}, |f(x)| \leq 1 \rbrace$.

Take $U = \lbrace f \in \mathcal{C}^0_{\rightarrow 0}(\mathbb{R}, \mathbb{R}) \mid f(x) < 1 + e^x \rbrace$. Then $U$ clearly contains $B$. Moreover, $U$ is open : indeed, if you take $f \in U$, then the function $g: x \mapsto 1+e^x-f(x)$ is positive everywhere, and tends to $1$ at $-\infty$ and to $+\infty$ at $+\infty$, so by continuity there exists $M>0$ such that $g(x) \geq M$ for every $x$. Then the open ball in $E$ centered at $f$ with radius $M/2$ is contained in $U$.

But $U$ does not contain any closed ball centered at $0$ with radius $>1$. Indeed, let $\varepsilon >0$. Let $x_0 \in \mathbb{R}$ such that $1+e^{x_0} < 1 + \varepsilon$. Then you can easily construct a function $f \in \mathcal{C}^0_{\rightarrow 0}(\mathbb{R}, \mathbb{R}) $, such that $f(x_0)=\max_{x \in \mathbb{R}} |f(x)|=1+\varepsilon$. Such a function belongs to the ball centered at $0$ with radius $1+ \varepsilon$, but does not belong to $U$.