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I was reading a proof that rewrote an inequality in the form:

$$b^Tx +x^T A x \le \alpha$$

for $b,x \in \mathbb{R}^n$ and $\alpha \in \mathbb{R}$, and with $A$ positive semidefinite. It then concluded that the solution set is convex. Why is this the case? I can see that $b^Tx$ is an affine function in $x$ and the latter is some sort of ellipsoid? But I'm not sure why their sum in the inequality would lead one to conclude so readily that the solution set is convex?

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1 Answer 1

up vote 3 down vote accepted

Note that both $b^Tx$ and $x^TAx$ are convex functions, and that the sum of convex functions is convex, thus $b^Tx+x^TAx$ is convex. It is a fact that the sublevel sets of a convex function $f$, i.e. the sets $\{x:f(x)\le \alpha\}$, are convex.

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Not quite. $x^T A x$ is convex only when matrix $A$ is positive definite / semidefinite. –  Rod Carvalho Sep 27 '12 at 4:32
    
@RodCarvalho Which the OP specifies $A$ is. –  Alex Becker Sep 27 '12 at 4:32
    
@RodCarvalho I hardly think it's necessary to make an answer self-contained; in fact, to do so I would need to include the question, which is hardly ever done. It is standard practice not to repeat the assumptions made in the OP. –  Alex Becker Sep 27 '12 at 4:36
    
Yes that makes sense! The only part I don't see that well is the convexity of the quadratic form with $A$ posdef. Is the Hessian some multiple of $A$? I'm not sure how to differentiate it without expanding the expression first. –  Palace Chan Sep 27 '12 at 13:48
    
@PalaceChan The Hessian is $A$. Note that $$x^TAx=\sum\limits_{i,j} a_{ij}x_ix_j$$ so the partial derivatives are just $a_{ij}$. –  Alex Becker Sep 27 '12 at 17:50

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