Proving convexity of $S = \{x: x^tQx \leq \beta\}$

Question

Let $Q \in \mathbb R^{n \times n}$ be symmetric positive semidefinite matrix, and $\beta > 0$. Show that $$S := \{x \in \mathbb R^n: x^tQx \leq \beta\}$$ is a convex set.

Take $x,y \in S, \lambda \in [0, 1]$.

$(\lambda x + (1-\lambda)y)^tQ(\lambda x + (1-\lambda)y) = \lambda^2x^tQx + (1-\lambda)^2y^tQy + \lambda(1-\lambda)x^tQy + \lambda (1-\lambda)y^tQx$

$y^tQx$ is a real number so it's trivially equal to his transpose, so $y^tQx = (y^tQx)^t = x^tQ^ty = x^tQy$ from symmetry of $Q$.

So we have $\lambda^2x^tQx + (1-\lambda)^2y^tQy + 2\lambda(1-\lambda)x^tQy \leq \lambda ^2 \beta+(1-\lambda)^2 \beta+2\lambda(1-\lambda)x^tQy =\\ 2\lambda^2\beta -2\lambda \beta+\beta + 2\lambda(1-\lambda)x^tQy$

$\lambda \in [0,1]$ so $\lambda ^2 \leq \lambda$, so $2\lambda^2\beta - 2\lambda \beta \leq 0$

Hence $ 2\lambda^2\beta -2\lambda \beta+\beta + 2\lambda(1-\lambda)x^tQy \leq \beta+2\lambda(1-\lambda)x^tQy$

I'm unsure how to continue now

Answer 1

Every positive semidefinite matrix $Q$ has a unique positive semidefinite square root. In fact, if you orthogonally diagonalise $Q$ as $UDU^T$, then $Q^{1/2}$ is just equal to $UD^{1/2}U^T$, where $D^{1/2}$ is the entrywise square root of the nonnegative diagonal matrix $D$.

It follows that the inequality $x^TQx\le\beta$ can be rewritten as $\|Q^{1/2}x\|\le\sqrt{\beta}$. You can now prove the convexity of $S$ by triangle inequality.

Answer 2

The mapping $x \mapsto x^TQx$ is convex (second-derivative test). A set defined by a convex function is convex: if $g$ is a convex function, then $$ \{x : \ g(x) \le 0\} $$ is a convex set by elementary computations.