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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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    This function is vector-valued, what do you mean by convexity in this case?2012-05-18
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    ... and it doesn't hold in 1d: $a\mapsto (a-b)^2$ is convex on $\mathbb R$, but $a \mapsto (a-b)^2a$ isn't.2012-05-18
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    I guess that $(a-b)'(a-b)$ means $(a-b)^T(a-b)$, which is the same thing as $\|a-b\|^2=\langle a-b,a-b \rangle$, where $\langle\cdot,\cdot\rangle$ is the standard [scalar product](http://en.wikipedia.org/wiki/Dot_product).2012-05-18
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    @martini: Sorry, the function was actually (a-b)'a'Sb(a-b) where the term in the middle a'Sb is a scalar (S is a square matrix and constant like b). Would this function be convex?2012-05-18
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    Not in general. If I didn't a computation error, if e. g. $b^TSb < 0$, $b \ne 0$, then $f''(b/2)[b,b] < 0$.2012-05-18
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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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