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I am trying to understand the following proof, which seems to be simple, but I think I am missing something. I couldn't find the open interval about $L(y)$.

Lemma 4.2: Let $A$ be an open convex set and $L : X\to \mathbb{R}$ be a nonzero linear continuous functional. Then $L[A]$ is open.

Proof: Pick $x_0$ with $L(x_0)\neq 0$. For any $y\in A$, $\{t\mid y+tx_0\in A\}$ is an open interval about $0$, so $L[A]$ contains an open interval about $L(y)$.

Obs.: $X$ is a locally convex topological vector space, but you can take it Banach if you want.

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  • $\begingroup$ $X$ does not need to be Banach (as my answer makes clear), but if it is at least an F-space, we also have an easier proof via the open mapping theorem. Is the open mapping theorem available to you at this point in the textbook? $\endgroup$ – Michael Lee Jul 29 '17 at 3:45
  • $\begingroup$ @MichaelLee Yes, it is! =D $\endgroup$ – Filburt Jul 29 '17 at 18:14
  • $\begingroup$ Great, I've added a section to my answer on how to show this using the open mapping theorem when $X$ is an F-space. $\endgroup$ – Michael Lee Jul 29 '17 at 18:45
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As $A$ is open, for any $y\in A$ there is an open neighborhood $U\ni y$ such that $U\subseteq A$. As $X$ is a locally convex topological vector space, it has a base of translated balanced, absorbent, convex sets, so without loss of generality we can take $U$ to be a balanced, absorbent, convex set that has been translated by $y$. We let $U' = U-y$ and note that we cannot have $L(U') = \{0\}$; as $U'$ is absorbent and $L(tU') = \{0\}$ for all $t > 0$, this would imply that $L(X) = \{0\}$, which contradicts that $L$ is nonzero. Thus, we let $x'\in U'$ such that $L(x')\neq 0$. As $U$ is a translate of a balanced set, $y+tx'\in U$ for all $\lvert t\rvert\leq 1$. Therefore, $$L(y)+tL(x')\in L(U)\subseteq L(A)$$ for all $t\in [-1, 1]$, so there is an interval of radius $\lvert L(x')\rvert$ around $L(y)$ in $L(A)$.


Note that in the case that $X$ is an F-space (which includes the case that $X$ is a Banach space), this also follows from the open mapping theorem. From Rudin's Functional Analysis:

Open Mapping Theorem: For continuous linear $L : X\to Y$, where $X$ is an F-space and $Y$ is a topological vector space, either $L(X)$ is meager in $Y$ or (1) $L(X) = Y$, (2) $L$ is an open mapping, and (3) $Y$ is an F-space.

As $L : X\to \mathbb{R}$ is nonzero and linear, it must be surjective, as for $L(x)\neq 0$ and $y\in \mathbb{R}$, we have that $L(cx) = y$ for $c = \frac{y}{L(x)}$. As $L(X) = \mathbb{R}$ is nonmeager in $\mathbb{R}$, $L$ must be an open mapping (and therefore $L(A)$ is open).

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