Fix a subgradient $p$ at $x$ and let $d_{1}(h),...,d_{n}(h)$ be defined as in the statement of Lemma 3.2. Let

$$I=\min\left\{i:\frac{d_{i}(h)}{h^{1/2}}\rightarrow 0\text{ as }h\rightarrow 0\right\}.$$

Fix $\delta$ small. Then we can find a sequence $h_{k}\rightarrow 0$ and $\eta$ depending only on $p$ such that

$$d_{I}(h_{k})<\delta h_{k}^{1/2},$$

and $d_{i}(h_{k})>\eta h_{k}^{1/2}$ for all $i<I$. Rotate the axes so that the $e_{i}$ are the axes for the John ellipsoid of $S_{h_{k},p}^{v}(x)$ and assume by translation that $x=0$.

Take the restriction of $v$ to the subspace spanned by $e_{I},...,e_{n}$, and call this restriction $w$. Let

$$S_{k}^{w}=S_{h_{k},p}^{v}(x)\cap\{x_{1}=...=x_{I-1}=0\},$$

the slice of the section $S_{h_{k},p}^{v}(x)$ in this subspace. Then since

$$d_{1}(h_{k})d_{2}(h_{k})...d_{n}(h_{k})\leq Ch_{k}^{\frac{n+1}{2}}$$

and $v$ grows at most quadratically in the first $I-1$ directions, we have

$$|S_{k}^{w}|_{\mathcal{H}^{n-I+1}}\leq\frac{C}{\eta^{(I-1)/2}}h_{k}^{\frac{n+2-I}{2}}.$$

Using this and Lemma 2.4,

$$Mw(S_{k}^{w})\geq c\eta^{(I-1)/2}h_{k}^{\frac{n-I}{2}}.$$

Finally, let $r_{k}=C(n)d_{I}(h_{k})$, with $C(n)$ taken large enough that

$$S_{k}^{w}\subset B_{r_{k}/2}(x).$$

By strict quadratic growth, $\nabla v(B_{r_{k}}(x))$ contains a ball of radius $r_{k}/2$ around every point in $\nabla v(S_{k}^{w})$. It follows that

	$\displaystyle Mv(B_{r_{k}}(x))$	$\displaystyle\geq c(n)Mw(S_{k}^{w})r_{k}^{I-1}$
		$\displaystyle\geq ch_{k}^{\frac{n-I}{2}}r_{k}^{I-1}$
		$\displaystyle\geq\frac{c}{\delta^{n-I}}r_{k}^{n-1}.$

By Lemma 3.2 we have $I\leq n-1$, so the conclusion follows. ∎