4
$\begingroup$

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?

1 Answers 1

4

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.

  • 0
    Not quite. $x^T A x$ is convex *only* when matrix $A$ is positive definite / semidefinite.2012-09-27
  • 0
    @RodCarvalho Which the OP specifies $A$ is.2012-09-27
  • 0
    @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.2012-09-27
  • 0
    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.2012-09-27
  • 0
    @PalaceChan The Hessian *is* $A$. Note that $$x^TAx=\sum\limits_{i,j} a_{ij}x_ix_j$$ so the partial derivatives are just $a_{ij}$.2012-09-27