# Approximation of closed convex set in separable Banach space

Let $V$ be a separable Banach space with $\{x_1, x_2, \ldots\}$ as its countable dense subset and $K\subset V$ be a closed, convex set. We define $$V_n = \text{span}\{x_1, x_2, \ldots, x_n\}, \quad \forall n\ge 1.$$

It is easy to prove that $\displaystyle \bigcup_{n=1}^\infty V_n$ is dense in $V$.

How can we construct the sets $K_n$ such that $$K_n\subset V_n, K_n \text{ is convex and } \bigcup_{n=1}^\infty K_n \text{ is dense in } K?$$

Thank you very much.

• $K_n:=V_n\cap K$? – Bananach Sep 4 '17 at 7:13
• @Bananach: This does not work. Let $V = \mathbb R$, $\{x_i\}$ be the rational numbers and $K = \{\pi\}$. – gerw Sep 4 '17 at 9:05
• @gerw I didn't check if it works, but I don't understand your example: wouldnt $V_n=\mathbb{R}$ and $K_n=\{\pi\}$? – Bananach Sep 4 '17 at 14:16
• @Bananach: Of course. I missed the span... – gerw Sep 4 '17 at 19:46
• Please specify what you mean by "dense in $K$" here. If you expect $K_n\subset K$, the answer is "can't be done". If $K_n\subset K$ is not required, then let $K_n=V_n$. – user357151 Sep 5 '17 at 22:35

Have you tried answering your questions? The first one is obvious because each x_n is in the union. For the second, take K_n to be the set of those points of K intersected with V_n whose norms do not exceed n.

• This does not work. Let $V = \mathbb R$, $\{x_i\}$ be the rational numbers and $K = \{\pi\}$. – gerw Sep 4 '17 at 9:06
• Are thinking of the span of $x_i$'s as the set of rational linear combinations? If $V=R$ then each $V_n$ is V. – Kabo Murphy Sep 15 '17 at 11:27

Hint: Take a countable subset $\{y_i\}$ of $K$, approximate each of these points $y_i$ by points in $V_n$ and take the convex hull.

• Could you be more specific please? – Tien Kha Pham Sep 4 '17 at 10:34
• Did you try to follow my hint? At which step are you struggling? I will not give you a step-by-step solution. – gerw Sep 4 '17 at 11:12
• I'm considering whether the set $\{y_n\}$ is necessarily dense in $K$? Following your hint, I approximate each $y_i$ by a sequence $x_{i, n}$ such that $x_{i, n}\in V_n$ and $x_{i, n}\to y_i$. For each $n$, I define $S_n$ is the convex hull of $\{x_{i, n}: i = 1, 2, \ldots\}$, and set $K_n = S_n \cap K$. Howerver, it seems that defining $K_n$ in this way does not work. – Tien Kha Pham Sep 4 '17 at 12:04
• Oh, you want to have $K_n \subset K \cap V_n$? Then it cannot work since $K \cap V_n$ might be empty. – gerw Sep 4 '17 at 19:48

It is not always possible to construct such $K_n.$

For example consider the case where $\displaystyle \bigcup_{n=1}^\infty V_n \neq V$ then pick $x \in V \setminus \displaystyle \bigcup_{n=1}^\infty V_n$ and define $K = \{x\}.$