Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let Q $\in \mathbb{R}^{d \times d} $ and A $\in$ $\mathbb{R}^{d' \times d} $ be two matrices. Let b $\in \mathbb{R}^d$ and c $\in \mathbb{R}^{d'}$ be two vectors. Suppose that d' < d.

I want to consider the quadratic programming problem for x $\in \mathbb{R}^d$:
minimize f(x):= $\frac{1}{2}x\cdot Qx -b\cdot x \ \ $ subject to Ax = c

Prove that a local min is also a global min, by as follows:

Let $x_0$ be a local min of f(x). Argue by contradiction and suppose that there exists a point $x_1$ such that f($x_1$) < f($x_0$)

a) Consider the function $\phi$(s):= $\frac{1}{2}x(s)\cdot Qx(s) -b\cdot x(s) \ \ $ where x(s):=(1-s)$x_0$ +s$x_1$. Prove that $\phi'$(0)=0 and $ \phi''$(0) $\geq$ 0

b) Compute $\phi$'(0) and $\phi''$(0), and show that $\phi(s)$=$\phi(0)$ + s$\phi$'(0) + $\frac{s^2}{2} \phi''(0)$

c) Prove that $\phi(s) \geq f(x_0)$ for all s $\geq$ 0

How does this conclude that a local min is also a global min?


So for a) x(0)= $x_0$ which implies that $\phi(0)$ = f($x_0$). Since $x_0$ is a local minimum, f'(0) = 0 and f''(0) $\geq$ 0.

for c) the statement is just a result from b).

I just dont know how to find the derivatives of $\phi(s)$ for b). Need some help.

share|cite|improve this question

So we want to take derivative of the following quadratic form with respect to $s$: $$ \phi(s) = \frac{1}{2} \big((1-s)x_0 + sx_1\big)^T Q \big((1-s)x_0 + sx_1\big) - b\cdot \big((1-s)x_0 + sx_1\big). $$ Back to $f(x)$, we know that: $$ \frac{d}{dx} f(x) = \frac{d}{dx} (\frac{1}{2} x^T Qx - b\cdot x)= \frac{1}{2}(Q+Q^T)x + b, $$ hence by chain rule: $$ \begin{aligned} \phi'(s) &= \left(\frac{1}{2}(Q+Q^T)x + b\right)\Bigg|_{x = (1-s)x_0 + sx_1} \cdot ((1-s)x_0 + sx_1)' \\ &= \frac{1}{2}(Q+Q^T)((1-s)x_0 + sx_1)\cdot (x_1 - x_0) + b\cdot (x_1 - x_0) \\ &=\frac{1}{2}(x_1 - x_0)^T(Q+Q^T)((1-s)x_0 + sx_1) + (x_1 - x_0)^T\, b . \end{aligned} $$ Then we can see that if $Q = Q^T$, and $x_0$ solves $Q x_0 =b $, then $\phi'(0) = 0$.

share|cite|improve this answer

Hint: Use the chain rule and the product rule and also that $x'(s) = x_1 - x_0$.

share|cite|improve this answer
Could you explain where/how I would use the chain rule and product rule? thanks – sarah Jun 1 '13 at 1:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.