$\{x^TQx \leq \alpha\}$ is convex when $Q$ is SPD

Question

Suppose $Q^T = Q \in \mathbb R^{n \times n}$ and $x^T Q x \geq 0$ for all $x \in \mathbb R^n$. Show the set $$S = \{ x \in \mathbb R^n \mid x^T Q x \leq \alpha\}$$ where $\alpha > 0$ is given, is convex.

My attempt:

Let $x,y \in S$. We want to show that $tx + (1-t)y \in S$ where $t \in [0, 1]$

If we show $\langle Q(tx + (1-t)y), tx + (1-t)y\rangle \leq \alpha $ we are done.

Simplying this, we get

$t^2\langle Qx, x \rangle + (1-t)^2\langle Qy, y\rangle + 2t(1-t)\langle Qx,y \rangle \leq t^2\alpha+(1-t)^2\alpha + 2t(1-t)\langle Qx,y\rangle$

Now, we can define $f(t) = \alpha t^2 + \alpha (1-t)^2$, where $t \in [0, 1]$, $\alpha > 0$. It is very easy to check that the maximum of this function is actually $\alpha$. So we can simplify even further, and say

$t^2\alpha+(1-t)^2\alpha + 2t(1-t)\langle Qx,y\rangle \leq \alpha + 2t(1-t)\langle Qx, y\rangle$

I'm not sure where to go from here.

Answer 1

Hint: use the fact that $$ 2\langle Qx,y\rangle\le\langle Qx,x\rangle+\langle Qy,y\rangle. $$

Answer 2

Maybe an easier approach is to write $Q=L^TL$ for some $L\in \mathbb{R}^{n\times n}$ by PSDness, so your set is just $\{x:\|Lx\|_2^2\leq \alpha\}=\{x:\|Lx\|_2\leq \sqrt{\alpha}\}$. Can you show this set is convex for any $\alpha$?