= Solution
A <symplectic capacity> assigns $c(X,\omega)\in[0,\infty]$ to <symplectic manifolds> in a fixed dimension, is monotone under <symplectic embeddings>, satisfies $c(X,a\omega)=a\,c(X,\omega)$ for $a>0$, and is nontrivial in the sense that
$$
0<c(B^{2n}(1))\leq c(B^2(1)\times\mathbb R^{2n-2})<\infty.
$$
The second domain is the standard <symplectic cylinder>. A <normalized symplectic capacity> assigns $\pi$ to both unit domains, hence $\pi R^2$ to both radius-$R$ domains. The general axioms already give invariance under <symplectomorphisms> and quadratic scaling under coordinate dilation. For the rigidity argument, use the assumed existence of capacities in each dimension, including after adding a symplectic plane; normalization is not needed.
Let $\phi_j$ be <symplectomorphisms> converging locally uniformly to a smooth <diffeomorphism> $\phi$. Work in <Darboux charts>. For a sufficiently small <solid ellipsoid> $E$ centered at a chosen point and $0<a<1$, uniform convergence and <degree of a continuous mapping> give, for large $j$,
$$
\phi_j(aE)\subseteq\phi(E)\subseteq\phi_j(a^{-1}E).
$$
For the second inclusion, $\phi^{-1}\phi_j$ is uniformly close to the identity on the larger ellipsoid. Its boundary stays outside $\overline E$ and is homotopic there to the identity, so it has degree one about every point of $E$ and must cover $E$. Monotonicity and scaling now give $a^2c(E)\leq c(\phi(E))\leq a^{-2}c(E)$. Letting $a$ tend to one proves <capacity preservation under uniform limits>.
At the chosen point, write $T=D\phi$. The rescaled maps $(\phi(tz)-\phi(0))/t$ preserve capacities of ellipsoids and converge uniformly on compact sets to $T$. Applying the same sandwich argument proves that $T$ preserves the capacities of all <solid ellipsoids>; so does $T^{-1}$.
We prove the required <linear capacity rigidity> explicitly. Write the standard <symplectic form> as $\Omega$. Suppose $\Omega(u,v)=1$ and
$$
0<|\Omega(T^Tu,T^Tv)|=\lambda^2<1.
$$
Complete $u,v$ to a <symplectic basis>, and complete $T^Tu/\lambda$, with the appropriately signed $T^Tv/\lambda$, to another <symplectic basis>. If their basis matrices are $P,P'$, then $A=(P')^{-1}T^TP$ has first two columns $\lambda e_1,\pm\lambda f_1$. Hence $A^T$ sends the unit ball into the <symplectic cylinder> of radius $\lambda$. This matrix differs from $T$ by symplectic linear maps, so it preserves capacities of ellipsoids. Its first coordinate pair is multiplied by $\lambda$, with a possible sign, and therefore $(A^T)^k$ sends the unit ball into the cylinder of radius $\lambda^k$. Monotonicity gives
$$
0<c(B^{2n}(1))\leq\lambda^{2k}c(Z^{2n}(1))
$$
for every $k$, contradicting finiteness of the right-hand unit-cylinder capacity as $k$ tends to infinity. For a normalized capacity, the first iterate already gives the contradiction.
A zero pairing is handled by a small perturbation of $u,v$. A pairing of absolute value greater than one gives the same contradiction for $T^{-1}$, after normalizing the transformed pair. Thus
$$
|\Omega(T^Tu,T^Tv)|=|\Omega(u,v)|
$$
for every pair. The squared bilinear forms agree, so their difference times their sum is the zero polynomial. The real polynomial ring is an <integral domain>; consequently one factor vanishes identically. Therefore $T$ is either symplectic or <anti-symplectic>.
To eliminate the negative sign, repeat the argument for $\phi_j\times\operatorname{id}_{\mathbb R^2}$, which are <symplectomorphisms> for $\omega\oplus\omega_{\mathbb R^2}$. If $\phi^*\omega=-\omega$ at the chosen point, the limit pulls this product form back to $-\omega\oplus\omega_{\mathbb R^2}$, which is neither the product form nor its negative. This contradicts the preceding derivative classification. Thus $\phi^*\omega=\omega$ everywhere. We have proved the <Eliashberg–Gromov rigidity theorem>: \b[the symplectomorphism group is closed in the $C^0$ topology on diffeomorphisms].
Finally put $E=W^\perp$, with perpendicularity taken for the standard Euclidean <inner product>. Its dimension is two, so being nonisotropic means that the restricted <symplectic form> is <nondegenerate>. For the standard compatible complex structure $J$, $W^\omega=JE$. If $Jv\in W\cap JE$, then $v\in E$ and $\omega(v,e)=0$ for every $e\in E$, forcing $v=0$. Thus $W$ is a <symplectic subspace> and admits a <symplectic basis> extended to the whole space.
In the resulting linear symplectic coordinates, $W$ is $\{q_1=p_1=0\}$. The projection of the bounded set $U$ to the $(q_1,p_1)$ plane is bounded, so $U+W$ lies in a <symplectic cylinder> of some finite radius $R$. As $U$ is nonempty and open, it contains a ball of some radius $r>0$, which is also contained in $U+W$. Monotonicity and scaling give
$$
\boxed{0<r^2c(B^{2n}(1))\leq c(U+W)\leq R^2c(Z^{2n}(1))<\infty},
$$
as in <capacity of a bounded set thickened by a symplectic subspace>. For a <normalized symplectic capacity>, the two bounds simplify to $\pi r^2$ and $\pi R^2$.
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