Let $Q : E \to F$ be a quadratic form induced by a symmetric bilinear form $B : E \times E \to F$ defined in a finite dimensional real normed vector space $E$, with values in the normed vector space $F \supseteq E$ (continuous inclusion). I already know that the image $C= Q(E)$ is a cone in $F$. How do I prove that it is also closed? Also, is it true that the convex hull of $C$ is closed?
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$\begingroup$ Really? Do you have a counterexample in mind? What if the norms are induced by an inner product? $\endgroup$– Eduardo LongaCommented May 25, 2022 at 0:08
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$\begingroup$ I'll post a counterexample. But in fact, I'm quite confused about the convexity now. (Probably because I should actually sleep rather than doing maths...) $\endgroup$– Jochen GlueckCommented May 25, 2022 at 0:11
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$\begingroup$ May I ask whether you found out whether $Q(E)$ is always convex? $\endgroup$– Jochen GlueckCommented May 26, 2022 at 17:28
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$\begingroup$ I didn't found out :( but I suspect it may not be always convex $\endgroup$– Eduardo LongaCommented Jun 2, 2022 at 20:32
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1$\begingroup$ Today I happened to find a simple counterexample which shows that $Q(E)$ is not always convex: let $E = \mathbb{R}^2$ and $F = \mathbb{R}^{2 \times 2}$, and let $B(x,y) = xy^{\operatorname{T}}$ for all $x,y \in E$. Then the range of $Q$ is the set of the symmetric matrices of rank $\le 1$, which is not convex. $\endgroup$– Jochen GlueckCommented Sep 12, 2022 at 11:54
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1 Answer
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In general, $Q(E)$ is not closed.
Counterexample. Let $E = F = \mathbb{R}^2$ and set $$ B(x,y) := \begin{pmatrix} x_1y_1 \\ x_1y_2 + x_2y_1 \end{pmatrix}. $$ Then $$ Q(x) := \begin{pmatrix} x_1^2 \\ 2x_1x_2 \end{pmatrix}, $$ so $Q(E)$ is the open right half plane together with the origin.
answered May 25, 2022 at 0:16