Does the accessibility lemma from convex analysis also hold in infinite dimensions?

Question

Accessibility lemma. Let $C \subset \mathbb R^n$ be convex and nonempty. For all $\lambda \in [0, 1)$, $x \in \text{ri}(C)$ and $y \in \overline{C}$ we have $(1 - \lambda) x + \lambda y \in \text{ri}(C)$, where $$ \text{ri}(C) := \{ x \in C: \exists \varepsilon > 0: B(x, \varepsilon) \cap \overline{\text{aff}}(C) \subset C \} $$ is the relative interior of $C$ with respect to $\overline{\text{aff}}(C)$ and $\text{aff}(C)$ is the affine hull of $C$, and $B(x, \varepsilon) := \{ y \in \mathbb R^n: | x - y |_2 < \varepsilon \}$ is the Euclidean norm ball. (The name is taken from Exercise 4 in Chapter 3.1 of Niculescu, Persson - Convex Functions and Their Applications.)

Note that since $\mathbb R^n$ is finite-dimensional, $\overline{\text{aff}}(C) := \overline{\text{aff}(C)} = \text{aff}(C)$.

I propose the following proof for this statement when $\mathbb R^n$ is replaced by any normed space $(E, \| \cdot \|)$, which uses the proof in Rockafellar's "Convex Analysis" (1970, Princeton Press) (Thm. 6.1, p. 45) but I don't know if it is correct.

Proof. Let $x \in \text{ri}(C)$, $y \in \overline{C}$ and $\lambda \in [0, 1)$. We want to show that $(1 - \lambda) x + \lambda y \in \text{ri}(C)$.
As $x \in \text{ri}(C)$, there exists a $\varepsilon > 0$ such that \begin{equation*} B_{\varepsilon}(x) \cap \overline{\text{aff}}(C) \subset C. \end{equation*} Since $y \in \overline{C}$, we have $y \in C + B_{\rho}(0)$ for every $\rho > 0$. Hence for all $\rho > 0$ with $\rho < \varepsilon \frac{1 - \lambda}{1 + \lambda}$ we have (see also this answer) \begin{align*} B_{\varepsilon}\big((1 - \lambda) x + \lambda y\big) & = (1 - \lambda) x + \lambda y + B_{\varepsilon}(0) \\ & \subset (1 - \lambda) x + \lambda \big( C + B_{\rho}(0) \big) + B_{\varepsilon}(0) \\ & = (1 - \lambda) \left[x + \frac{1 + \lambda}{1 - \lambda} B_{\rho}(0) \right] + \lambda C \\ & = (1 - \lambda) B_{\rho \cdot \frac{1 + \lambda}{1 - \lambda}}(x) + \lambda C \\ & \subset (1 - \lambda) B_{\varepsilon}(x) + \lambda C \end{align*} and thus \begin{align*} B_{\varepsilon}\big((1 - \lambda) x + \lambda y\big) \cap \overline{\text{aff}}(C) & \subset (1 - \lambda) B_{\varepsilon}(x) \cap \overline{\text{aff}}(C) + (\lambda C) \cap \overline{\text{aff}}(C) \\ & \overset{(\star)}{=} (1 - \lambda) \left[B_{\varepsilon}(x) \cap \overline{\text{aff}}(C)\right] + (\lambda C) \cap \overline{\text{aff}}(C) \\ & \subset (1 - \lambda) C + (\lambda C) \cap \overline{\text{aff}}(C) = (1 - \lambda) C + \lambda C = C, \end{align*} where in the last step we use that $C$ is convex and in the third last step that $\lambda C \subset C \subset \overline{\text{aff}}(C)$.

(We also use that $(A + B) \cap C = (A \cap C) + (B \cap C)$ and that $A \subset B$ implies $(A \cap C) \subset (B \cap C)$ as well as $\lambda (A \cap B) = (\lambda A) \cap (\lambda B)$ for sets $A, B, C \subset E$ and $\lambda \ne 0$ as well as the fact that (closed) affine hull commutes with linear transformations such as scalings.)

Any comments on the correctness of this proof are highly appreciated. I am not sure about the equality $(\star)$: I think that since $T(z) := (1 - \lambda) z$ is a linear map, $T(\text{aff}(A)) = \text{aff}(T(A))$ for any set $A$, so that $(1 - \lambda) \text{aff}(C) = \text{aff}((1 - \lambda) C)$. Since $\lambda \in (0, 1)$, we have $(1 - \lambda) C \subset C$ and thus $\text{aff}\big((1 - \lambda) C\big) \subset \text{aff}(C)$. Applying the closure should yield $(1 - \lambda) \overline{\text{aff}}(C) \subset \overline{\text{aff}}(C)$. Since $\lambda (A \cap B) = (\lambda A) \cap (\lambda B)$ for any non-zero scalars $\lambda$ and sets $A, B \subset E$, we still need the other inclusion which I am not sure how to prove (or if it even holds). In my finite-dimensional intuition, this certainly makes sense, since it doesn't matter if we take the scaled ball and intersect it with an affine subspace to yield a scaled disk, or first intersecting the ball with the affine subspace and then scale it.

Answer 1

The proof in the question is specious at a couple of places.

We also use that $(A + B) \cap C = (A \cap C) + (B \cap C)$

That is incorrect. For example, $A=\{1\}$, $B=\{2\}$ and $C=[3,5]\subset\Bbb R^1$. Then $A+B=\{3\}$, $(A+B)\cap C=\{3\}$. However, both $A\cap C$ and $B\cap C$ are empty.

The question/answer linked is not related at all. Note that "$(A+B)$ is the symmetric difference and has the form of $(A \cup B) \backslash (A \cap B)$" over there.

Since $\lambda \in (0, 1)$, we have $(1 - \lambda) C \subset C$.

That is incorrect. For example, let $C$ be the interval $[1,2]\subset\Bbb R^1$. For $\lambda = \frac23\in(0,1)$, $\frac32\in C$, we have $(1-\lambda)\frac32=\frac12\notin C$.

---

However, what you want to prove, the accessibility lemma when $\Bbb R^n$ is replaced by any normed space is correct.

Here is a proof. The basic idea is a comparison between points $x,y,w$ and $x',y',w'$, where the variables will be explained soon.

Let $E$ be a normed vector space. Let $C \subset E$ be convex and nonempty. Let $\lambda \in [0, 1)$, $x \in \text{ri}(C)$ and $y \in \overline{C}$. We want to show that $(1 - \lambda) x + \lambda y \in \text{ri}(C)$.

Since the case when $\lambda=0$ is obviously true, assume $\lambda\not=0$.

Let $w= (1 - \lambda) x + \lambda y$.
As $x \in \text{ri}(C)$, there exists an $\varepsilon > 0$ such that $$ B_\varepsilon(x)\cap\overline{\text{aff}}(C)\subset C$$
Since $y\in \overline C$, there exists $y'\in C$ such that $\|y'-y\|<\frac{(1-\lambda)\varepsilon}{3\lambda}$.
Consider an arbitrary point $w'\in B_{(1-\lambda)\varepsilon/3}(w)$.
Let $x'\in E$ satisfy $w'=(1-\lambda) x'+\lambda y'$, i.e., $x'=\frac{w'-\lambda y'}{1-\lambda}$.
Then $x'-x=\frac{(w'-w) - \lambda(y'-y)}{1-\lambda}$ $$\begin{aligned} \|x'-x\|&\le\frac{\|w'-w\|+\lambda\|y'-y\|}{1-\lambda}\\ &\le \frac{(1-\lambda)\varepsilon/3+\lambda\frac{(1-\lambda)\varepsilon}{3\lambda}}{1-\lambda}\\ &=\frac23\varepsilon\\ &<\varepsilon\\ \end{aligned}$$

Suppose $w'\in\overline{\text{aff}}(C)$. Since $x', y', w'$ are colinear where $y', w'$ are in $\overline{\text{aff}}(C)$, $x'$ is in $\overline{\text{aff}}(C)$. The inequality above implies $x'\in C$. Hence $w'\in C$ since $w'$ is a convex combination of $x',y'\in C$. We have shown that $B_{(1-\lambda)\varepsilon/3}(w)\cap\overline{\text{aff}}(C)\subset C$, i.e., $w\in\text{ri}(C)$.