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?



Image of a quadratic form is a closed cone
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.


