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I take this example from here. However, I would like to fill in the detail to make sure I do not miss something important. Could you have a check on my attempt?

Let $X := L_2([0,1])$ and $S=\{f \in X \mid f\ge 0 \text{ a.e.}\}$. Then $S$ is convex, has empty interior, and whose affine hull is $X$.

Proof:

First, we show that $\operatorname{int} S = \emptyset$. Fix $f \in S$ and $r>0$. Because the space $C_b([0, 1])$ of continuous bounded function on $[0, 1]$ is dense in $X$, there is $g \in C_b([0, 1])$ such that $\|f-g\|_2 < r/2$. Assume $g$ is bounded by $\alpha>0$. We pick $x_0 \in (0, 1]$ and $\beta<0$ such that $x_0 (\alpha-\beta) < r/2$. We define $h := 1_{[0, x_0]} \beta + 1_{(x_0, 1]} g$. Then $h \in X \setminus S$. On the other hand, $\|h-f\|_2 \le \|h-g\|_2 + \|g-f\|_2 \le r/2+r/2 = r$. This implies $S$ has no interior points.

If $f \in X$, then $f = 1 \cdot f^+ + (-1) \cdot f^- + 1 \cdot 0$ where $f^+=\max\{f, 0\}, f^-= \max\{-f, 0\}$. Obviously, $f^+, f^- , 0\in S$ and $1+(-1)+1 = 1$. As such, $\operatorname{aff} S= X$.

Analyst
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    Empty interior could be obtained slightly simpler. Let $f\in S.$ Define $$f_n(x)=\begin{cases} -1& 0 \le x\le {1\over n}\ f(x) & {1\over n}<x\le 1\end{cases}$$ Then $f_n$ tends to $f$ in $L^2$-norm. – Ryszard Szwarc Jun 08 '22 at 17:17
  • @RyszardSzwarc could you post your comment as an answer so that I can accept it? – Analyst Jun 09 '22 at 16:07

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