[proofplan] The Gromov width is defined by testing how large a standard symplectic ball can be symplectically embedded into a given symplectic manifold. Monotonicity follows by composing a ball embedding with the given symplectic embedding. Conformality follows from the coordinate dilation of the standard ball, whose pullback multiplies $\omega_0$ by the square of the dilation factor. The two normalisations come from the evident inclusions for the lower bounds, the symplectic volume obstruction for the ball upper bound, and Gromov non-squeezing for the cylinder upper bound. [/proofplan]

[step:Prove monotonicity by composing symplectic embeddings]Let \begin{align*} F:(M,\omega)\to(N,\eta) \end{align*} be a symplectic embedding. Fix $r>0$ such that there exists a symplectic embedding \begin{align*} \varphi:(B^{2n}(r),\omega_0)\to(M,\omega). \end{align*} Then the composition \begin{align*} F\circ\varphi:B^{2n}(r)\to N \end{align*} is a smooth embedding. Moreover, \begin{align*} (F\circ\varphi)^*\eta=\varphi^*(F^*\eta)=\varphi^*\omega=\omega_0. \end{align*} Thus $F\circ\varphi$ is a symplectic embedding of $(B^{2n}(r),\omega_0)$ into $(N,\eta)$. Therefore every number $\pi r^2$ admitted in the defining supremum of $c_G(M,\omega)$ is also admitted in the defining supremum of $c_G(N,\eta)$. Taking suprema gives \begin{align*} c_G(M,\omega)\le c_G(N,\eta). \end{align*}[/step]

[guided]The monotonicity axiom says that a symplectic embedding cannot decrease the set of standard balls that can be fitted into the target. Let \begin{align*} F:(M,\omega)\to(N,\eta) \end{align*} be a symplectic embedding. To compare the two suprema, take an arbitrary admissible radius for $(M,\omega)$. Thus let $r>0$ and suppose there is a symplectic embedding \begin{align*} \varphi:(B^{2n}(r),\omega_0)\to(M,\omega). \end{align*} The natural map into $N$ is the composition \begin{align*} F\circ\varphi:B^{2n}(r)\to N. \end{align*} It is a smooth embedding because it is a composition of smooth embeddings. It is symplectic because pullbacks compose contravariantly: \begin{align*} (F\circ\varphi)^*\eta=\varphi^*(F^*\eta). \end{align*} Since $F$ is symplectic, $F^*\eta=\omega$, and since $\varphi$ is symplectic, $\varphi^*\omega=\omega_0$. Hence \begin{align*} (F\circ\varphi)^*\eta=\omega_0. \end{align*} So the same radius $r$ that contributes $\pi r^2$ to $c_G(M,\omega)$ also contributes $\pi r^2$ to $c_G(N,\eta)$. Since this holds for every admissible $r$, the set whose supremum defines $c_G(M,\omega)$ is contained in the set whose supremum defines $c_G(N,\eta)$. Therefore \begin{align*} c_G(M,\omega)\le c_G(N,\eta). \end{align*}[/guided]

[step:Use coordinate dilations to prove positive conformality] Fix $\lambda>0$. For $a>0$, define the smooth dilation map \begin{align*} D_a:\mathbb R^{2n}\to\mathbb R^{2n} \end{align*} by \begin{align*} D_a(x)=ax. \end{align*} Since each coordinate function is multiplied by $a$, one has \begin{align*} D_a^*\omega_0=a^2\omega_0. \end{align*} Suppose $r>0$ and \begin{align*} \varphi:(B^{2n}(r),\omega_0)\to(M,\omega) \end{align*} is a symplectic embedding. Define the smooth map \begin{align*} \psi:B^{2n}(\sqrt{\lambda}r)\to M \end{align*} by \begin{align*} \psi=\varphi\circ D_{1/\sqrt{\lambda}}. \end{align*} Because $D_{1/\sqrt{\lambda}}$ maps $B^{2n}(\sqrt{\lambda}r)$ diffeomorphically onto $B^{2n}(r)$, $\psi$ is a smooth embedding. Furthermore, \begin{align*} \psi^*(\lambda\omega)=\lambda D_{1/\sqrt{\lambda}}^*(\varphi^*\omega)=\lambda D_{1/\sqrt{\lambda}}^*\omega_0=\lambda\cdot \frac{1}{\lambda}\omega_0=\omega_0. \end{align*} Thus \begin{align*} c_G(M,\lambda\omega)\ge \lambda c_G(M,\omega). \end{align*} Applying the same argument to the symplectic form $\lambda\omega$ and the scalar $1/\lambda$ gives \begin{align*} c_G(M,\omega)=c_G(M,(1/\lambda)(\lambda\omega))\ge \frac{1}{\lambda}c_G(M,\lambda\omega). \end{align*} Multiplying by $\lambda$ gives \begin{align*} \lambda c_G(M,\omega)\ge c_G(M,\lambda\omega). \end{align*} Combining the two inequalities yields \begin{align*} c_G(M,\lambda\omega)=\lambda c_G(M,\omega). \end{align*} [/step]

[step:Normalize the Gromov width of the standard ball] Fix $R>0$. The identity inclusion \begin{align*} \operatorname{id}_{B^{2n}(R)}:(B^{2n}(R),\omega_0)\to(B^{2n}(R),\omega_0) \end{align*} is a symplectic embedding, so \begin{align*} c_G(B^{2n}(R),\omega_0)\ge \pi R^2. \end{align*} For the reverse inequality, let $r>0$ and suppose \begin{align*} \varphi:(B^{2n}(r),\omega_0)\to(B^{2n}(R),\omega_0) \end{align*} is a symplectic embedding. Because $\varphi$ is symplectic, $\varphi^*\omega_0=\omega_0$, and therefore \begin{align*} \varphi^*\left(\frac{1}{n!}\omega_0^n\right)=\frac{1}{n!}\omega_0^n. \end{align*} Let $\mathcal L^{2n}$ denote [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb R^{2n}$. With the orientation determined by $\omega_0^n/n!$, the standard [symplectic volume form](/theorems/10035) on $\mathbb R^{2n}$ agrees with $\mathcal L^{2n}$. Since $\varphi$ is a smooth embedding between manifolds of the same dimension, the [inverse function theorem](/theorems/51) gives that $\varphi$ is a local diffeomorphism, and the embedding condition gives that $\varphi$ is a diffeomorphism from $B^{2n}(r)$ onto its image. Hence the full-dimensional embedded submanifold $\varphi(B^{2n}(r))\subset B^{2n}(R)$ is open in $B^{2n}(R)$. In particular, the image is $\mathcal L^{2n}$-measurable. Applying the change-of-variables theorem to this diffeomorphism and using $\varphi^*(\omega_0^n/n!)=\omega_0^n/n!$ gives \begin{align*} \mathcal L^{2n}(B^{2n}(r))=\mathcal L^{2n}(\varphi(B^{2n}(r))). \end{align*} Because $\varphi(B^{2n}(r))\subset B^{2n}(R)$, monotonicity of Lebesgue measure gives \begin{align*} \mathcal L^{2n}(\varphi(B^{2n}(r)))\le \mathcal L^{2n}(B^{2n}(R)). \end{align*} Using the Euclidean ball volume formula \begin{align*} \mathcal L^{2n}(B^{2n}(\rho))=\frac{\pi^n\rho^{2n}}{n!}, \end{align*} we obtain \begin{align*} \frac{\pi^n r^{2n}}{n!}\le \frac{\pi^n R^{2n}}{n!}. \end{align*} Hence $r\le R$, and therefore $\pi r^2\le \pi R^2$. Taking the supremum over all admissible $r$ gives \begin{align*} c_G(B^{2n}(R),\omega_0)\le \pi R^2. \end{align*} Thus \begin{align*} c_G(B^{2n}(R),\omega_0)=\pi R^2. \end{align*} Here the volume-preservation input is the standard symplectic volume obstruction for symplectic embeddings. [/step]

[step:Normalize the Gromov width of the standard cylinder] Fix $R>0$. Since \begin{align*} B^{2n}(R)\subset Z^{2n}(R), \end{align*} the inclusion \begin{align*} \iota:(B^{2n}(R),\omega_0)\to(Z^{2n}(R),\omega_0) \end{align*} is a symplectic embedding. Hence \begin{align*} c_G(Z^{2n}(R),\omega_0)\ge \pi R^2. \end{align*} For the reverse inequality, first suppose $n\ge 2$. If $r>0$ and there is a symplectic embedding \begin{align*} \varphi:(B^{2n}(r),\omega_0)\to(Z^{2n}(R),\omega_0), \end{align*} then the hypotheses of the [Gromov Non-Squeezing Theorem for Standard Symplectic Balls and Cylinders](/theorems/10075) [citetheorem:10075] are satisfied: $n\ge 2$, $r>0$, $R>0$, the domain is the standard symplectic ball $(B^{2n}(r),\omega_0)$, the target is the standard symplectic cylinder $(Z^{2n}(R),\omega_0)$, and $\varphi$ is the required symplectic embedding. The theorem therefore gives \begin{align*} r\le R. \end{align*} Therefore $\pi r^2\le \pi R^2$ for every admissible $r$, and so \begin{align*} c_G(Z^{2n}(R),\omega_0)\le \pi R^2. \end{align*} If $n=1$, then \begin{align*} Z^2(R)=B^2(R), \end{align*} because both are the set of points $(q_1,p_1)\in\mathbb R^2$ satisfying $q_1^2+p_1^2<R^2$. The ball normalization already proved gives \begin{align*} c_G(Z^2(R),\omega_0)=\pi R^2. \end{align*} Combining the lower bound with the upper bound in all dimensions gives \begin{align*} c_G(Z^{2n}(R),\omega_0)=\pi R^2. \end{align*} [/step]

[step:Conclude that the Gromov width satisfies the capacity axioms] The first step proves monotonicity under symplectic embeddings. The second step proves conformality for every positive scalar $\lambda>0$. The third and fourth steps prove the ball-cylinder normalization: \begin{align*} c_G(B^{2n}(R),\omega_0)=c_G(Z^{2n}(R),\omega_0)=\pi R^2. \end{align*} Therefore $c_G$ satisfies monotonicity, positive conformality, and the standard ball-cylinder normalization on the class of $2n$-dimensional symplectic manifolds. Hence the Gromov width is a symplectic capacity. [/step]