Is $f$ in the vector space of cubic spline functions? Let $S_{X,3}$ be the vector space of cubic spline functions on $[-1,1]$ in respect to the points $$X=\left \{x_0=-1, x_1=-\frac{1}{2}, x_2=0, x_3=\frac{1}{2}, x_4\right \}$$ I want to check if the function $$f(x)=\left ||x|^3-\left |x+\frac{1}{3}\right |^3\right |$$ is in $S_{X,3}$. 
We have that \begin{align*}f(x)&=\left ||x|^3-\left |x+\frac{1}{3}\right |^3\right |\\ & =\begin{cases}|x|^3-\left |x+\frac{1}{3}\right |^3 , & |x|^3-\left |x+\frac{1}{3}\right |^3\geq 0 \\-|x|^3+\left |x+\frac{1}{3}\right |^3 , & |x|^3-\left |x+\frac{1}{3}\right |^3<0\end{cases} \\ & =\begin{cases}|x|^3-\left |x+\frac{1}{3}\right |^3 , & |x|^3\geq \left |x+\frac{1}{3}\right |^3 \\-|x|^3+\left |x+\frac{1}{3}\right |^3 , & |x|^3<\left |x+\frac{1}{3}\right |^3\end{cases} \\ & = \begin{cases}|x|^3-\left |x+\frac{1}{3}\right |^3 , & |x|\geq \left |x+\frac{1}{3}\right | \\-|x|^3+\left |x+\frac{1}{3}\right |^3 , & |x|<\left |x+\frac{1}{3}\right |\end{cases}\end{align*} 
The function is piecewise a polynomial of degree smaller or equal to $3$, right? 
Now we have to check if $f$ is continuous on $[-1,1]$. 
How could we continue to get the definition of $f$ ? 

For $x\in(-1/3,0)$ we have
$$f(x)=\left|x^3+\left(x+\frac{1}{3}\right)^3\right|$$
which has strictly positive slope, because $x\mapsto x^3$ is monotonous with zero derivative only at $x=0$.
On $(-1/3,0)$ the first term goes from negative to zero, and the second term goes from zero to positive.
Hence the sum goes from negative to positive, and somewhere within $(-1/3,0)$ the sum is zero with positive slope. Taking the absolute value results in non-differntiability at such a point, so it is not a spline, because you have no knots in the interval $(-1/3,0)$.
