Since \(Q\) is positive semidefinite, the function \(f(\vec x)\) is a quadratic form (see the slides \(\vec x^T A \vec x\) to confirm this) which is convex. However, we are asked to prove it using the definition, not just citing properties of quadratic forms.
We have the following function: \[f(\vec x) := \vec x^\top Q \vec x, \qquad \vec x\in \mathbb{R}^n,\] where \(Q \in \mathbb{R}^{n \times n}\) is positive semidefinite and symmetric. That is: \[z^\top Q z \ge 0 \quad \text{for all } z.\]
We prove that \(f\) is convex on \(\mathbb{R}^n\) using the definition: \[f(\vec y+ \alpha(\vec x-\vec y)) - \alpha f(\vec x) - (1-\alpha) f(\vec y) \le \vec 0\] for all \(\vec x,\vec y\in \mathbb{R}^n\) and all \(\alpha \in [0,1]\).
For the first term on the left-hand side, we have \[\vec z:= \vec y+ \alpha(\vec x-\vec y) = (1-\alpha)\vec y+ \alpha \vec x.\] Then \[\begin{aligned}
f(\vec z)
&= \vec z^\top Q \vec z\\
&= \big((1-\alpha)\vec y+ \alpha \vec x\big)^\top
Q
\big((1-\alpha)\vec y+ \alpha \vec x\big).
\end{aligned}\] Using symmetry of \(Q\), \[f(z)
=
(1-\alpha)^2 \vec y^\top Q y
+ 2\alpha(1-\alpha)\vec x^\top Q \vec y
+ \alpha^2 \vec x^\top Q \vec x.
\]
For the second part of the left-hand side, the affine combination: \(\alpha f(\vec x)
+ (1-\alpha) f(\vec y)\), we have: \[\alpha f(\vec x) + (1-\alpha)f(\vec y)
=
\alpha \vec x^\top Q \vec x+ (1-\alpha)\vec y^\top Q \vec y.\]
Putting the terms together, we have: \[\begin{aligned}
&f(\vec z) - \alpha f(\vec x) - (1-\alpha)f(\vec y) \\
&=
\alpha^2 \vec x^\top Q \vec x- \alpha \vec x^\top Q \vec x
+ (1-\alpha)^2 \vec y^\top Q \vec y- (1-\alpha)\vec y^\top Q \vec y\\
&\quad + 2\alpha(1-\alpha)\vec x^\top Q \vec y.
\end{aligned}\]
Factoring terms yields \[\begin{aligned}
&=
-\alpha(1-\alpha)\vec x^\top Q \vec x
-\alpha(1-\alpha)\vec y^\top Q \vec y
+ 2\alpha(1-\alpha)\vec x^\top Q \vec y\\
&=
\alpha(1-\alpha)
\Big(
- \vec x^\top Q \vec x
- \vec y^\top Q \vec y
+ 2\vec x^\top Q \vec y
\Big).
\end{aligned}\]
Observe that \[\vec x^\top Q \vec x+ \vec y^\top Q \vec y- 2\vec x^\top Q \vec y
=
(\vec x-\vec y)^\top Q (\vec x-\vec y).\]
Therefore, \[f(\vec z) - \alpha f(\vec x) - (1-\alpha)f(\vec y)
=
-\alpha(1-\alpha)(\vec x-\vec y)^\top Q (\vec x-\vec y).\]
Since \(\alpha(1-\alpha) \ge 0\) for all \(\alpha \in [0,1]\) and \(Q\) is positive semidefinite, we have that \[(\vec x-\vec y)^\top Q (\vec x-\vec y) \ge 0,\] we conclude that \[f(\vec y+ \alpha(\vec x-\vec y)) - \alpha f(\vec x) - (1-\alpha) f(\vec y) \le 0.\]
Thus, \[f(\vec x) = \vec x^\top Q \vec x\] is convex on \(\mathbb{R}^n\). If \(Q\) is positive definite, the inequality is strict for \(\vec x\neq \vec y\) and \(\alpha \in (0,1)\), so \(f\) is strictly convex.
