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

Sorry.I don't know enough about convex optimization.
How can I know this function for the $\theta$ vector
is a convex function?
The other variables are arbitrary constants.
Thanks in advance.

share|cite|improve this question
If you will rewrite it with no brackets, you will realize, that there are term only of the order 1 and 2. That means that the function is quadratic. Maybe I just didn't understand your question? – Dmitry Laptev Jun 13 '12 at 14:34
I'm asking why its convex on theta @JohnSmith – cloudygoose Jun 13 '12 at 14:36
Oh, yeah I has a mistake in my question @JohnSmith – cloudygoose Jun 13 '12 at 14:39
I still can make no sense of the question (perhaps my bad, though): what is $\,\theta\,$ and where does it appear in the question? What are $\,x_1\,,\,x_2\,,\,y_0$? What's the domain and what the range of this assumed function?? – DonAntonio Jun 13 '12 at 14:41

The function $f(\theta) = \theta_0+\theta_1x_1+\theta_2x_2-y_0$ is linear, and the function $\phi(t) = t^2$ is convex. It is straightforward to show the composition is convex. Suppose $\lambda \in [0,1]$:

$$\phi(f(\lambda x +(1-\lambda)y)) = \phi(\lambda f(x)+(1-\lambda)f(y)) \leq \lambda \phi(f(x)) + (1-\lambda)\phi(f(y)).$$

share|cite|improve this answer

If a function $x \in \mathbb{R}^n \mapsto f(x) \in \mathbb{R}$ is convex then a function $y \in \mathbb{R}^m \mapsto f(Ay + b) = g(y)$ is also convex for any matrix $A \in \mathbb{R}^{n \times m}$ and any vector $b \in \mathbb{R}^n$ because $$ g(\alpha y' + (1-\alpha)y'') \equiv f(A(\alpha y' + (1-\alpha)y'') + b) = \\ = f(\alpha(Ay' + b) + (1-\alpha)(Ay''+b)) \leqslant \alpha f(Ay'+b)+(1-\alpha)f(Ay''+b) = \\ = \alpha g(y') + (1-\alpha)g(y''). $$ In your case function $f(x) = x^2$ is convex because $f''(x) = 2 > 0$, $A = (1,x_1,x_2)$, $b = -y_0$.

share|cite|improve this answer

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.