\begin{proposition}\label{prop:kkt1}
Let $\KKT_n$ be the problem: given $f\in C^1([0,1]^n)$ together with its gradient, find $x\in[0,1]^n$ with
$x=P_{[0,1]^n}(x-\nabla f(x))$, i.e.\ a Karush--Kuhn--Tucker point of $\min_{[0,1]^n}f$. Then
$\KKT_1\seqv\IVT$, and $\KKT_n\slev\BFT_n$ for all $n$.
\end{proposition}

\begin{proof}
$\KKT_n\slev\BFT_n$: the KKT points are the fixed points of the continuous self-map
$x\mapsto P_{[0,1]^n}(x-\nabla f(x))$ of the cube.
$\KKT_1\slev\IVT$: the function $G(x)=x-P_{[0,1]}(x-f'(x))$ is continuous with $G(0)\le0\le G(1)$, and its
zeros are the KKT points. Extend $G$ linearly to $[-1,2]$ with $G(-1)=-1$ and $G(2)=1$; the extension is negative
on $[-1,0)$, positive on $(1,2]$, and changes sign strictly, so $\IVT$ applies.
$\IVT\slev\KKT_1$: given $h$ with $h(0)<0<h(1)$, let $f(x)=\int_0^xh$. Then $0$ and $1$ are not KKT points
($f'(0)<0$, $f'(1)>0$), and the interior KKT points are the zeros of $h$.
\end{proof}

So in dimension one the exact KKT problem is as easy as finding a zero, and every computable instance has a
computable solution. Since $\BFT_n\seqv\WKLp$ for $n\ge2$ \cite{BrattkaLeRouxMillerPauly2019}, while gradient fields
are very special vector fields, it is natural to ask whether $\KKT_n$ is strictly below $\BFT_n$ in higher
dimensions, as $\CLS$ is (conjecturally) strictly below $\PPAD$ in the finite world. This is \cref{prob:kkt}.