Is f in the vector space of cubic spline functions?

[mathmari]

Hey!

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 $f$ ? (Wondering)

 

[I like Serena]

Yes. And those pieces should be between those $x_i$.
That is, between each $x_i$ and $x_{i+1}$ we should have a polynomial of degree $\le 3$.
It doesn't look like we will get that, since the derivative at $x=-\frac 13$ will be discontinuous. (Worried)

Now we have to check if $f$ is continuous on $[-1,1]$.

How could we continue to get the definition of $f$ ?

How about we start with the inner expression $|x|^3-\left |x+\frac{1}{3}\right|$ and expand it for the cases:
\begin{cases} -1 \le x<-\frac 13 \\ -\frac 13\le x < 0 \\ 0 \le x \le 1\end{cases}
Then all the absolute signs should disappear. (Thinking)