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I have a $f(\boldsymbol{x}) = \boldsymbol{x}^{\boldsymbol{T}} \boldsymbol{Q} \boldsymbol{x}$.

$\boldsymbol{Q}$ is an $n \times n$ symmetric positive semidefinite matrix. How can I show that $f(\boldsymbol{x})$ is convex on the domain $R^n$?

Attempt: By definition of convex, for any $x,y\in\mathbb R$, we have $$f(\frac{x+y}2)\leq\frac12(f(x)+f(y))$$ Thus it is sufficient to reduce and prove that $$\frac12(x+y)^TQ(x+y)\leq x^TQx+y^TQy\\ x^TQy+y^TQx\leq x^TQx+y^TQy$$ Namely $$(x-y)^TQ(x-y)\geq0$$

At this point I am stuck on how to use positive semi-definite - assuming my logic is sound up to this point.

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    What does positive semidefinite mean? – 2017-01-30
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    It means that the scalar $x^T Q x$ is guaranteed to be non-negative. I guess I'm stuck on what more I have to show. Do I just state the definition and end it there? – 2017-01-30
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    Is it correct to use the definition of convexity that I used? I think this is really where I am struggling. – 2017-01-30
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    If you have already proved that this definition of convexity is equivalent to the standard definition, or if you are allowed to assume this fact, then you can finish off your proof just by invoking the definition of positive semidefinite in the final step. – 2017-01-30
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    A different proof is to note that $f$ is differentiable on $\mathbb R^n$ and $\nabla^2 f(x) = 2Q$ is positive semidefinite for all $x$. Since the Hessian of $f$ is positive semidefinite for all $x$, it follows that $f$ is convex. – 2017-01-30
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    I see no reason whatsoever to use anything but the Hessian test here. Why make life harder than necessary? ;-) – 2017-01-30

1 Answers 1

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(a) For $\;f:\mathbb{R}\to \mathbb{R}$, $\;f(x)=x^2$ we verify that $f$ is convex. In fact, $$(\alpha x+(1-\alpha )y)^2\leq \alpha x^2+(1-\alpha)y^2 \Leftrightarrow\\ \alpha^2x^2+2\alpha (1-\alpha)xy+(1-\alpha)^2y^2-\alpha x^2-(1-\alpha)y^2\leq 0\Leftrightarrow\\(\alpha^2-\alpha)x^2+(\alpha^2-\alpha)y^2+ 2(\alpha^2-\alpha)xy\leq 0\Leftrightarrow\\ (\alpha^2-\alpha)(x+y)^2\leq 0.$$ The last inequality is verified because $\alpha^2-\alpha \leq 0$ for all $\alpha\in [0,1]$. That is, $$f(\alpha x+(1-\alpha )y)\leq \alpha f(x)+(1-\alpha)f(y)\quad \forall{x,y}\in{V},\; \forall{\alpha }\in{[0,1]}.$$ and $f$ is a convex function.

(b) For $\;f(\boldsymbol{x}) = \boldsymbol{x}^{\boldsymbol{T}} \boldsymbol{Q} \boldsymbol{x}\;$ with $\;\boldsymbol{Q}\;$ positive semidefinite, we can express $$f(\boldsymbol{x}) = \boldsymbol{x}^{\boldsymbol{T}} \boldsymbol{Q} \boldsymbol{x}=\boldsymbol{(\boldsymbol{P}x})^{\boldsymbol{T}} \boldsymbol{D} (\boldsymbol{P}\boldsymbol{x})=x_1^2+\cdots+x_r^2$$ and apply (a).