# Image of a quadratic form is a closed cone

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?

• Really? Do you have a counterexample in mind? What if the norms are induced by an inner product? May 25 at 0:08
• 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...) May 25 at 0:11
• May I ask whether you found out whether $Q(E)$ is always convex? May 26 at 17:28
• I didn't found out :( but I suspect it may not be always convex Jun 2 at 20:32
• Thanks for your response! Jun 2 at 20:38

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.