A multimodal quartic receives an exact spatial branch-and-bound gap certificate

The identity $f(x)=(x^2-\frac{1}{2}{)}^2-\frac{1}{4}$ shows $f(x)\ge -\frac{1}{4}$, with equality at $x=\pm \sqrt{1\mathrm{/}2}$. If $s=\sqrt{1\mathrm{/}2}$, then $s\ge 0$ and $s^2=1\mathrm{/}2$ by [F2], so $s>0$. We have $s<1$: otherwise either $s=1$, contradicting $s^2=1\mathrm{/}2$, or $1<s$, which by strict monotonicity of squaring on nonnegatives gives $1<s^2=1\mathrm{/}2$, also impossible. Hence both minimizers lie in $[-1,1]$, and they are distinct because $s>0$. Squaring is monotone on the nonnegatives

At the root box, the DAG product $t=x\times x$ encloses as $[-1,1]$ under ordinary interval multiplication; then $u=t\times t$ also encloses as $[-1,1]$, so $f=u-t$ encloses as $[-2,2]$. The root midpoint is $0$, with exact feasible value $f(0)=0$; its initial gap is $0-(-2)=2>9\mathrm{/}10$, so the root cannot yet be closed as a global certificate.

The root uses the first call. Bisect at zero and evaluate the lower child on the second call and the upper child on the third call, completing both reports within $N=3$. On each child $[-1,0]$ and $[0,1]$, ordinary interval multiplication gives $t=x\times x\in [0,1]$, then $u=t\times t\in [0,1]$, so $f=u-t$ has lower bound $-1$. The child midpoints are $-1\mathrm{/}2$ and $1\mathrm{/}2$, and both give $f=1\mathrm{/}16-1\mathrm{/}4=-3\mathrm{/}16$, so the incumbent becomes $U=-3\mathrm{/}16$ and the global lower bound is $L=-1$.

The gap is $U-L=13\mathrm{/}16<9\mathrm{/}10$. The feasible midpoint and the complete two-leaf ledger therefore certify epsilon-global optimality. Its actual objective is $1\mathrm{/}16$ above the minimum, within the declared tolerance.