[proofplan] The forward construction is the ordinary inclusion of the Euclidean ball into the cylinder when $r\le R$, and the only point to check is that this inclusion preserves the standard symplectic form. The converse is the deep part: we use Gromov's pseudo-holomorphic curve argument as an external analytic input, in its ball-cylinder area-obstruction form. Here $\pi$ denotes the usual circle constant. That input compactifies the cylinder in the first symplectic coordinate direction, produces a pseudo-holomorphic curve in the line class through a point of the embedded ball, and compares its total area $\pi R^2$ with the monotonicity lower bound $\pi r^2$ inside the ball. [/proofplan] [step:Construct the symplectic embedding when $r\le R$] Assume \begin{align*} r\le R. \end{align*} Define the map \begin{align*} \iota:B^{2n}(r)&\to \mathbb R^{2n} \end{align*} by \begin{align*} \iota(x)=x. \end{align*} For every $x=(q_1,p_1,\dots,q_n,p_n)\in B^{2n}(r)$, the Euclidean norm condition gives \begin{align*} q_1^2+p_1^2\le |x|^2<r^2\le R^2. \end{align*} Hence $\iota(x)\in Z^{2n}(R)$, so $\iota$ restricts to a smooth embedding \begin{align*} \iota:B^{2n}(r)\to Z^{2n}(R). \end{align*} Since $\iota$ is the identity map on the ambient coordinate functions, its pullback fixes each coordinate one-form: \begin{align*} \iota^*dq_i=dq_i \end{align*} and \begin{align*} \iota^*dp_i=dp_i \end{align*} for every $i\in\{1,\dots,n\}$. Therefore \begin{align*} \iota^*\omega_0 = \sum_{i=1}^n \iota^*(dq_i\wedge dp_i) = \sum_{i=1}^n dq_i\wedge dp_i = \omega_0. \end{align*} Thus a symplectic embedding exists when $r\le R$. [guided] Assume \begin{align*} r\le R. \end{align*} The natural candidate is the inclusion map, because the cylinder only restricts the first symplectic coordinate plane. Define \begin{align*} \iota:B^{2n}(r)&\to \mathbb R^{2n} \end{align*} by \begin{align*} \iota(x)=x. \end{align*} We must verify that the image actually lies in the cylinder. Let \begin{align*} x=(q_1,p_1,\dots,q_n,p_n)\in B^{2n}(r). \end{align*} By the definition of $B^{2n}(r)$, \begin{align*} |x|^2=q_1^2+p_1^2+\cdots+q_n^2+p_n^2<r^2. \end{align*} Since every omitted term in the sum is nonnegative, we have \begin{align*} q_1^2+p_1^2\le |x|^2<r^2\le R^2. \end{align*} This is exactly the defining inequality for membership in $Z^{2n}(R)$, so $\iota$ is a map \begin{align*} \iota:B^{2n}(r)\to Z^{2n}(R). \end{align*} It is a smooth embedding because it is the restriction of the identity map on $\mathbb R^{2n}$. It remains to check the symplectic condition. The standard symplectic form is \begin{align*} \omega_0=\sum_{i=1}^n dq_i\wedge dp_i. \end{align*} Since $\iota$ is the identity on all coordinate functions, pullback by $\iota$ leaves each coordinate one-form unchanged: \begin{align*} \iota^*dq_i=dq_i \end{align*} and \begin{align*} \iota^*dp_i=dp_i. \end{align*} Using linearity of pullback and compatibility of pullback with wedge products, \begin{align*} \iota^*\omega_0 = \sum_{i=1}^n \iota^*(dq_i\wedge dp_i) = \sum_{i=1}^n \iota^*dq_i\wedge \iota^*dp_i = \sum_{i=1}^n dq_i\wedge dp_i = \omega_0. \end{align*} Therefore the inclusion is the required smooth symplectic embedding. [/guided] [/step] [step:Invoke Gromov's pseudo-holomorphic curve obstruction for the converse] We use the following deep external result from Gromov's pseudo-holomorphic curve proof of non-squeezing, recorded in the ball-cylinder normalisation part of [citetheorem:10076]: if $n\ge 2$, $r>0$, $R>0$, and there exists a smooth embedding \begin{align*} \varphi:B^{2n}(r)\to Z^{2n}(R) \end{align*} such that \begin{align*} \varphi^*(\omega_0|_{Z^{2n}(R)})=\omega_0|_{B^{2n}(r)}, \end{align*} then \begin{align*} \pi r^2\le \pi R^2. \end{align*} For context, this result is proved by compactifying the cylinder in the $(q_1,p_1)$-direction, choosing an almost complex structure compatible with the compactified symplectic form and standard on the embedded ball, producing a pseudo-holomorphic curve in the line class through a point of $\varphi(B^{2n}(r))$, and applying the monotonicity theorem for pseudo-holomorphic curves to obtain the lower area bound $\pi r^2$ inside the embedded ball while the line class has total area $\pi R^2$. Now assume that a smooth embedding \begin{align*} \varphi:B^{2n}(r)\to Z^{2n}(R) \end{align*} satisfies \begin{align*} \varphi^*(\omega_0|_{Z^{2n}(R)})=\omega_0|_{B^{2n}(r)}. \end{align*} The hypotheses of the cited obstruction are satisfied by the present $n$, $r$, $R$, and $\varphi$. Hence \begin{align*} \pi r^2\le \pi R^2. \end{align*} Since $\pi>0$, division by $\pi$ gives \begin{align*} r^2\le R^2. \end{align*} Because $r>0$ and $R>0$, taking positive square roots yields \begin{align*} r\le R. \end{align*} Thus existence of a symplectic embedding implies $r\le R$. [guided] We now isolate exactly what is being used for the hard direction. The cited ball-cylinder obstruction in [citetheorem:10076] says that, for $n\ge 2$ and positive radii $r$ and $R$, a smooth symplectic embedding \begin{align*} \varphi:B^{2n}(r)\to Z^{2n}(R) \end{align*} forces the area inequality \begin{align*} \pi r^2\le \pi R^2. \end{align*} The role of the pseudo-holomorphic curve theory is precisely to prove this obstruction: after compactifying the cylinder in the first symplectic coordinate plane, one produces a pseudo-holomorphic curve in the line class through a point of the embedded ball. Monotonicity gives at least $\pi r^2$ area inside the image of the ball, while the line class has total area $\pi R^2$. Assume now that such a smooth embedding $\varphi:B^{2n}(r)\to Z^{2n}(R)$ exists and satisfies \begin{align*} \varphi^*(\omega_0|_{Z^{2n}(R)})=\omega_0|_{B^{2n}(r)}. \end{align*} The hypotheses of the cited obstruction are exactly the present hypotheses: $n\ge 2$, $r>0$, $R>0$, and $\varphi$ is a smooth embedding preserving the standard symplectic form. Therefore \begin{align*} \pi r^2\le \pi R^2. \end{align*} Since $\pi$ is positive, dividing both sides by $\pi$ gives \begin{align*} r^2\le R^2. \end{align*} Finally, the radii are positive [real numbers](/page/Real%20Numbers), so the positive square-root function preserves the inequality and yields \begin{align*} r\le R. \end{align*} Thus any smooth symplectic embedding of the ball into the cylinder forces the radius bound $r\le R$. [/guided] [/step] [step:Combine the two implications] The first step proves that if $r\le R$, then the standard inclusion gives a smooth embedding \begin{align*} B^{2n}(r)\to Z^{2n}(R) \end{align*} whose pullback sends $\omega_0|_{Z^{2n}(R)}$ to $\omega_0|_{B^{2n}(r)}$. The second step proves that any smooth embedding preserving these restricted standard symplectic forms forces $r\le R$. Therefore such a smooth symplectic embedding exists if and only if \begin{align*} r\le R. \end{align*} This proves the theorem. [/step]