Quite rare kind of proof of convexity for a quadratic function!

Question

Excuse me all of you in advance. I got this problem as an assignment but I am not really good doing proofs! If $f(x)=\frac12x^TQx+b^Tx+a$ is quadratic in $n$ variables, where $Q$ is symmetric. Show that if $f(x)$ is convex in $∣∣x∣∣<ε$, then $f(x)$ is convex in $R^n$. I have to get the gradient of $f(x)$ and then the Hessian. So the gradient is $\frac12(Q+Q^T)x+b$ and the Hessian would be $\frac12(Q+Q^T)$. But I need to get the eigenvalues of the Hessian to proof the convexity of $f(x)$ and I have no idea how to get this done. Any help would be appreciated.

Answer 1

Partial Answer: Given $\|x\|,\|y\| < \epsilon$ and $t\in [0,1],$ we have $$ F(tx + (1-t)y) \leq t F(x) + (1-t) F(y), $$ Expand $F(tx + (1-t)y)$ to get \begin{align*} F(tx + (1-t)y) &= \frac{1}{2} t^2 x^T Q x + t(1-t) x^T Q y + \frac{1}{2} (1-t)^2 y^T Q y + b^T(tx + (1-t)y) + a \\ &= F(tx) + F((1-t)y) + t(1-t) x^T Q y. \end{align*} Thus for $\|x\|,\|y\|<\epsilon$ and $t\in[0,1]$ we have $$ t(1-t) x^T Q y \geq 0 $$ To get convexity for arbitrary values in $\mathbb{R}^n,$ simply scale $x,y$ by some $\alpha >0$ and we see that $$ t(1-t) \alpha^2 x^T Q y \geq 0. $$

Answer 2

You just need to show that the eigenvalues are non negative.

Since $x \mapsto -b^Tx -a$ is convex, if $f$ is convex on $B(0,\epsilon)$ then so is $q(x) = f(x)-b^Tx -a = {1 \over 2} x^TQx$.

$Q$ is real and symmetric so has a basis of eigenvectors corresponding to real eigenvalues. Let $\lambda$ be an eigenvalue, and $v$ a corresponding eigenvector with $\|v\| \in (0,\epsilon)$. Then $q(tv) = {1 \over 2} t^2 \lambda \|v\|^2$. Since $q$ is convex on $B(0,\epsilon)$, we must have $\lambda \ge 0$ (otherwise look at the values of $q(\pm v)$ and $q(0)$ to obtain a contradiction). Hence all eigenvalues of $Q$ are non negative, and so $Q$ is positive semi definite.