Local critical-value lowering preserves the upper sublevel

The Morse lemma supplies the displayed coordinates after shrinking $\epsilon >0$. Such cutoffs exist: take a smooth function $0\le \eta <1$ with support compactly inside $(0,2\epsilon )$ and integral greater than $\epsilon$, and put $\mu (t)={\int }_t^{2\epsilon }\eta (s)ds$. A bump equal to a constant less than one on a sufficiently long closed subinterval gives $\eta$. The perturbation has support compactly inside the chart, so gluing by zero is smooth.

Put $x=\mathrm{\mid }u{\mathrm{\mid }}^2$, $y=\mathrm{\mid }v{\mathrm{\mid }}^2$. Then $dF=-2(1+{\mu }^{\mathrm{\prime }})u\cdot du+2(1-2{\mu }^{\mathrm{\prime }})v\cdot dv$. Both scalar magnitudes are positive, so its only chart critical point is $(0,0)$; its Hessian there has the same index, including empty coordinate blocks. Its value is $c-\mu (0)<c-\epsilon$. All other critical points and their values are unchanged.

Since $F\le f$, one inclusion of upper sublevels holds. Wherever $F\mathrm{\ne }f$, $x+2y<2\epsilon$, hence $f=c-x+y<c+\epsilon$; elsewhere the two functions coincide. This proves the reverse inclusion. If $F\ge c-\epsilon$, then $f\ge F\ge c-\epsilon$. Thus the modified closed band is a closed subset of the original compact band. Its only candidate critical point has been lowered out of it.