Proving that f(x)=||x-a|| is a convex function

Question

f(x)=||x-a||, where a=(0,1)^T

Can someone show why f(x) is convex?

Answer 1

We have $\renewcommand{\vec}{\mathbf}$ for $\vec x, \vec y \in X$ and $\tau \in [0,1]$ that

\begin{align} f((1-\tau)\vec x + \tau \vec y) &= \|(1-\tau) \vec x + \tau \vec y - \vec a \| \\ &= \| (1-\tau) (\vec x - \vec a) + \tau (\vec y - \vec a)\| \\ &\leq \|(1-\tau)(\vec x - \vec a)\| + \|\tau(\vec y - \vec a)\| \\ &= (1-\tau) \|\vec x - \vec a\| + \tau \| \vec y - \vec a \| \\ &= (1-\tau) f(\vec x) + \tau f(\vec y). \end{align}