Find the primitives of $f(x)$.

Question

Let $f\colon \mathbb{R}\to\mathbb{R}$, $$ f(x)=\left\{\begin{array}{l}\underset{t\leq x}{\inf}(t^2-t+1),\;x\leq\frac12\\\underset{t\geqslant x}{\sup}(-t^2+t+1),x>\frac12\end{array}\right.$$ Say if $f(x)$ has any primitives on $\mathbb{R}$ and if it has, write the primitive. I have no idea how to approach this problem since I have no idea how to proceed with the $\inf$ and $\sup$ functions and what they really are.

Answer 1

So we have these functions if $g(x) = x^2-x+1$ and $h(x)=-x^2+x+1$, both defined w.r.t $f(x)$.

Now, how do I really get the $inf$ and $sup$ in function of $x$?

Answer 2

As: $$t^2-t+1=\left(t -\frac{1}{2}\right)^2+1-\frac{1}{4}$$ so if $t \leq x \leq 1/2$: $$\left(t -\frac{1}{2}\right)^2 \leq \left(x -\frac{1}{2}\right)^2$$ and: $$\inf_{t \leq x} \left(t -\frac{1}{2}\right)^2+\frac{3}{4}=\left(x -\frac{1}{2}\right)^2+\frac{3}{4}$$ you can do the same for the $\sup$ to show that if $x >1/2$: $$\sup_{t \geq x} \left(-\left(t -\frac{1}{2}\right)^2+\frac{5}{4}\right)=-\left(x -\frac{1}{2}\right)^2+\frac{5}{4}$$ and then you can take the primitive for $x>1/2$ and for $x \leq 1/2$ and adjust the constants.

---

As pointed out by @HenningMakholm if $f$ as a jump at $x=1/2$ there is no continuous function $F$ such that $F'(x)=f(x)$ for all $x\in \Bbb R$.