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Eduardo Longa
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Eduardo Longa
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Image of a quadratic form is a closed and convex 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 andhull of $C$ is closed?

Image of a quadratic form is a closed and convex 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 convex and closed?

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?

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Eduardo Longa
  • 2.1k
  • 12
  • 11

Image of a quadratic form is a closed and convex 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 convex and closed?