Transfer of the analytic gradient inequality to a smooth metric

In positive dimension, at every point of an analytic chart the real-analytic-germ supplier gives a holomorphic complexification $H$ on a complex polydisc, with real restriction $f=U\circ {\phi }^{-1}$. By the several-variable smoothness corollary, $H$ has continuous real coordinate derivatives of every order. Holding the imaginary coordinates zero in each coordinate difference quotient shows inductively that every iterated real partial derivative of $f$ is the restriction of the corresponding real partial derivative of $H$. Those restrictions are continuous, so $f$ is smooth. Thus $U$ is smooth in the given smooth atlas, and its Riemannian gradient is defined. In dimension zero every point has a singleton chart and $U$ is locally constant, hence smooth as well.

In positive dimension apply the local analytic inequality to $f$ at $\phi (p)$, obtaining $c_0>0$ and ${\rho }_0\in [0,1)$. Choose an analytic coordinate ball centered there with compact closure inside both the chart image and the inequality neighborhood. By continuity of $f$, shrink the ball so $\mathrm{\mid }f-f(\phi (p))\mathrm{\mid }\le 1$. Put $\rho =(1+{\rho }_0)\mathrm{/}2$, so $0<\rho <1$ and $\rho \ge {\rho }_0$.

In positive dimension $m=\dim M$, fix $x$ in the inverse image of the chosen coordinate ball and put $y=\phi (x)$. By the coordinate basis theorem write any $v\in T_{x}M$ uniquely as $v={\sum }_{i<m}{a}_i{\mathrm{\partial }}_i{\mathrm{\mid }}_x$, with coordinate vector $a\in {\mathbb{R}}^m$. The differential and coordinate-derivation definitions, together with linearity of a tangent derivation, give $d{U}_x(v)=v([U])={\sum }_{i<m}{a}_i{\mathrm{\partial }}_{i}f(y)=⟨\mathrm{\nabla }f(y),a⟩$. This derives the coordinate pairing without identifying a covector with a vector by notation alone.

For $0<a\le 1$, logarithm monotonicity gives $\log a\le \log 1=0$. Multiplying by $\rho -{\rho }_0\ge 0$ yields $\rho \log a\le {\rho }_0\log a$, so exponential monotonicity and the power definition give $a^{\rho }=\exp (\rho \log a)\le \exp ({\rho }_0\log a)=a^{{\rho }_0}$. Apply this at $a=\mathrm{\mid }f-f(\phi (p))\mathrm{\mid }>0$ to the inequality of step 2.1. At $a=0$ the desired right side is zero because $\rho >0$, so the inequality follows from nonnegativity of the norm, with no evaluation of $0^0$. Consequently $\mathrm{\parallel }\mathrm{\nabla }f{\mathrm{\parallel }}_2\ge c_0\mathrm{\mid }U-U(p){\mathrm{\mid }}^{\rho }$ on the ball.

On the compact chart closure, choose $C>0$ from metric comparison, so $g_x(v,v)\le C\mathrm{\parallel }a{\mathrm{\parallel }}_2^2$ for the coordinates of step 2.2. If $\mathrm{\nabla }f(y)\mathrm{\ne }0$, choose $a=\mathrm{\nabla }f(y)\mathrm{/}\mathrm{\parallel }\mathrm{\nabla }f(y){\mathrm{\parallel }}_2$ and $v={\sum }_{i<m}{a}_i{\mathrm{\partial }}_i{\mathrm{\mid }}_x$. The Euclidean inner-product formula gives $\mathrm{\parallel }a{\mathrm{\parallel }}_2=1$ and $d{U}_x(v)=\mathrm{\parallel }\mathrm{\nabla }f(y){\mathrm{\parallel }}_2$. The metric-dual identity and Cauchy–Schwarz give $\mathrm{\parallel }\mathrm{\nabla }f(y){\mathrm{\parallel }}_2=\mathrm{\mid }{g}_x({\mathrm{grad}}_{g}U(x),v)\mathrm{\mid }\le \mathrm{\parallel }{\mathrm{grad}}_{g}U(x){\mathrm{\parallel }}_g\sqrt{g_x(v,v)}\le \sqrt{C}\mathrm{\parallel }{\mathrm{grad}}_{g}U(x){\mathrm{\parallel }}_g$. The same bound holds trivially if $\mathrm{\nabla }f(y)=0$. Combining with step 3.1 proves the claim on the inverse image $W$ of the ball with $c=c_0\mathrm{/}\sqrt{C}$. In dimension zero choose a singleton chart: $U$ is constant there and both sides vanish for any $c>0$, $0<\rho <1$.