Is this $f(x)$ convex? I want to check if the function $f(x)$ is convex, where
$$ f(x)=\bigl|a|x|-x\bigr|^2.$$
There are several possibilities to check if the function is convex:


*

*The second derivative: Not possible, because it is only subdifferential.

*Geometry: too complicated for this case.

*Convex inequality: $$f(tx+(1-t)y) \leqslant tf(x) + (1-t)f(y).$$


To 3: How to use the inequality? Can somebody help to put the function into the convex inequality?
 A: $f$ can be written as
$$
f(x) = \begin{cases} (a-1)^2 x^2 & x\ge0,\\
(-a-1)x^2 & x\le 0.
\end{cases}
$$
The first derivative is given by
$$
f(x) = \begin{cases}2 (a-1)^2 x & x\ge0,\\
2(-a-1)^2x & x\le 0,
\end{cases}
$$
which is a monotonically increasing function, hence $f$ is convex.
A: first of all you don't need the outer abs |.| because of the square. Hence
$$
f(x)=\vert a\vert x\vert-x\vert^2=(a\vert x\vert -x)^2=a^2\vert x\vert^2-2a\vert x\vert x +x^2
$$
Observe that $\vert x\vert^2=x^2$ for all $x\in \mathbb{R}$. Thus,
$$
f(x)=a^2 x^2-2a\vert x\vert x +x^2=(a^2+1)x^2-2a\vert x\vert x
$$
Now, we can consider $f$ for $x\geq0$ and $x<0$, i.e.
$$
f(x)=\begin{cases}
(a^2-2a+1)x^2,&\quad x\geq 0\\
(a^2+2a+1)x^2,&\quad x< 0
\end{cases}=\begin{cases}
(a-1)^2x^2,&\quad x\geq 0\\
(a+1)^2x^2,&\quad x< 0
\end{cases}
$$
In both cases you have a parabola for $a\in \mathbb{R}\setminus \{-1,+1\}$ which are opened up. Hence $f$ is convex. Otherwise one part of $f$ is 0, the other part a parabola open up. So, $f$ is again convex.
