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I know that the function $(\mathbf{a}-\mathbf{b})'(\mathbf{a}-\mathbf{b})$ is convex in $\mathbf{a}$ ($\mathbf{a}$ and $\mathbf{b}$ are vectors, not scalars). Would $(\mathbf{a}-\mathbf{b})'(\mathbf{a}-\mathbf{b})\mathbf{a}'$ which is a cubic function be convex too? Is there a simple way to check this?

Edit (after martini's comment): the function is $(\mathbf{a}-\mathbf{b})'\mathbf{a}'S \mathbf{b}(\mathbf{a}-\mathbf{b})$ where the term in the middle $\mathbf{a}'S\mathbf{b}$ is a scalar (S is a square matrix and constant like $\mathbf{b}$).

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    @martini: if S is positive semi-definite then b'Sb >=0. In such a case, would the function be convex?2012-05-18

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No. It is still false in one dimension (if it is true for vectors, it has to be true for scalars too).

Let $S = [1]$ the $1\times 1$ positive definite square matrix. Let $b = [1]$ the $1\times 1$ vector. Then your function reduces to

$ a\mapsto (a-1)^2 a = a^3 - 2a^2 + a $

Its second derivative is

$ 6a - 4 $

which is negative for all $a < 2/3$ and so the function is certainly not convex.