Minimizing a convex cost function I'm reviewing basic techniques in optimization and I'm stuck on the following. We aim to minimize the cost function
$$f(x_1,x_2) = \frac{1}{2n} \sum_{k=1}^n \left(\cos\left(\frac{\pi k}{n}\right) x_1 + \sin\left(\frac{\pi k}{n}\right) x_2\right)^2.$$
I'd like to show some basic properties, specifically what its Lipschitz constant is, whether or not it is strongly convex, and where it obtains its minimum.
In finding the Lipschitz constant, 
$$\frac{\partial f}{\partial x_1} = \frac{1}{n} \sum_{k=1}^n \left(\cos\left(\frac{\pi k}{n}\right) x_1 + \sin\left(\frac{\pi k}{n}\right) x_2\right)\cos\left(\frac{\pi k}{n}\right),$$$$\frac{\partial f}{\partial x_2} = \frac{1}{n} \sum_{k=1}^n \left(\cos\left(\frac{\pi k}{n}\right) x_1 + \sin\left(\frac{\pi k}{n}\right) x_2\right)\sin\left(\frac{\pi k}{n}\right),$$
but for arbitrarily large $x_1$ and $x_2$, doesn't this imply this imply the derivatives are unbounded and thus the function not Lipschitz?
I'm having a similar problem in computing the minimum of the function, as I set one of the above equations to zero, solve for $x_1$, plug it into the other, and solving for $x_2$ simply gets $x_2 (\cdots) = 0$, where $(\cdots)$ is some jarble of product sums but nevertheless a constant, and thus meaningless in determine the value of $x_2$.
 A: Looking at your derivatives is quite interesting since
$$\sum _{k=1}^n \cos\left(\frac{\pi  k}{n}\right) \cos \left(\frac{\pi 
   k}{n}\right) =\frac{n} {2}$$
$$\sum _{k=1}^n \cos\left(\frac{\pi  k}{n}\right) \sin \left(\frac{\pi 
   k}{n}\right) =0$$
$$\sum _{k=1}^n \sin\left(\frac{\pi  k}{n}\right) \sin \left(\frac{\pi 
   k}{n}\right) =\frac{n} {2}$$ This then leads to
$$\frac{\partial f}{\partial x_1} = \frac{x_1}{2} $$
$$\frac{\partial f}{\partial x_2} = \frac{x_2}{2} $$ So, the derivatives are zero for $x_1=0$ and $x_2=0$ and this is the only solution.
A: In response to your comment: a general useful fact is that any energy of the form
$$\sum_i \left(\sum_j \alpha_{i,j} x_j\right)^2$$
for constants $\alpha_{i,j}$ is convex. This is because this energy can be written in matrix form as
$$\|Ax\|^2 = x^T A^TA x$$
for $A_{i,j} = \alpha_{i,j}$; the Hessian of this energy is $2A^TA$ which is necessarily symmetric positive-semidefinite.
It follows that the energy is strongly convex if $A$ has full rank, which is the case for your objective function for $n>1$.
