Convexity properties of $ \frac{1}{2\pi} \cdot \int_{S_1(0)} \Vert x -y \Vert d \mathbb{\lambda}^1(y) $ Let $S_1(0)$ be the unit sphere in $\mathbb{R}^2$. For a point $x \in B_1(0)$ define the function $f(x)$ to be $$ \frac{1}{2\pi} \cdot \int_{S_1(0)} \Vert x -y  \Vert d \mathbb{\lambda}^1(y) $$
i.e for a point $x$ we are computing the distance to every point on the circle and integrating. The function is obviously rotational symmetric and has a minimum at $0$. What can be said about convexity of this function? 
I have tried to compute the function explicitly, therefore I have tried to solve the integral 
$$ f(t)= \int_0^{2 \pi} \sqrt{1+t^2 - 2t \cos(x)} dx $$
for values $t \in [0,1]$. This is the value of the function at a point x=(0,t) and by symmetry therefore the value of any point with $\Vert x \Vert =t$. I have not been able to solve this integral. Does there exist an analytical solution? 
 A: Rewriting $$1+t^2-2t\cos(x)=(1-t)^2+2t\big(1-cos(x)\big)=(1-t)^2+4t^2 \sin^2\big(\frac x 2\big)$$ we then find that $$\int \sqrt{1+t^2-2t\cos(x)}\,dx=2 (1-t)\, E\left(\frac{x}{2}|-\frac{4 t}{(1-t)^2}\right)$$ where appears the elliptic integral of the second kind.
Then $$f(t)=\int_0^{2\pi} \sqrt{1+t^2-2t\cos(x)}\,dx=4 (1-t) E\left(-\frac{4 t}{(1-t)^2}\right)$$
A: There is no need to deal with elliptic integrals to prove the convexity of our function. 
For any $z$ in the unit disk, let 
$$ f(z) = \frac{1}{2\pi}\int_{S^1} \|z-x\|\,dx \tag{1}$$
the average distance from the unit circle. Quite obviously, $f(z)$ is a continuous function, hence to prove its convexity it is enough to prove its midpoint-convexity, but
$$ f(z_1)+f(z_2)\geq 2\,f\left(\frac{z_1+z_2}{2}\right) \tag{2}$$
just follows from the definition $(1)$ and the triangle inequality in the form
$$ \|a\|+\|b\|\geq 2 \left\|\frac{a+b}{2}\right\|.\tag{3}$$

As an alternative, the analytic function defined on $(-1,1)$ by
$$ g(t) = \frac{1}{2\pi}\int_{0}^{2\pi}\sqrt{1+t^2-2t\cos x}\,dx $$
has a remarkable property: it is an even function and every coefficient of the Taylor series at $t=0$ is non-negative. Convexity is a straightforward consequence.
$$ g(t) = 1+\frac{t^2}{4}+\frac{t^4}{64}+\frac{t^6}{256}+\ldots \tag{4} $$
Since $f(z)$ is a radial function, we may use $(4)$ to provide lower bounds ($\lambda=\frac{1}{2}$) for the hessian matrix eigenvalues, too.
