Given the Fock space: $$\mathcal{F}(\mathcal{h}):=\bigoplus_0^\infty\mathcal{h}^{n}\text{ with } \mathcal{h}^{n}:=\bigotimes_1^n \mathcal{h},\mathcal{h}^0:=\mathbb{C}$$
Define the creation and annihilation operators by: $$a(f)\alpha_1\otimes\ldots\otimes\alpha_n=\sqrt{n}\langle f,\alpha_n\rangle\alpha_1\otimes\ldots\otimes\alpha_{n-1}\qquad f,\alpha_i\in\mathcal{h}$$ $$a(f)^\dagger\alpha_1\otimes\ldots\otimes\alpha_n=\sqrt{n}\langle f,\alpha_n\rangle\alpha_1\otimes\ldots\otimes\alpha_{n}\otimes f\qquad f,\alpha_i\in\mathcal{h}$$
How do I prove that: $$\|a(f)\psi\|\leq \sqrt{n}\|\psi\|\quad\psi\in\mathcal{H}^{n}$$ $$\|a(f)^\dagger\psi\|=\sqrt{n+1}\|\psi\|\quad\psi\in\mathcal{H}^{n}$$
For the annihilation operators an elementary calculation gives: $$\|a^*(f)\psi_n\|_{n+1}=\|\sqrt{n+1}f\otimes\psi_n\|_{n+1}=\sqrt{n+1}\|f\|\cdot\|\psi_n\|_n\quad\psi_n\in\mathcal{h}^n$$ Moreover it holds: $$\|a^*(f)\lambda\Omega\|_1=\|\lambda f\|_1=\sqrt{0+1}\|f\|\cdot\|\lambda\Omega\|_0\quad\lambda\Omega\in\mathcal{h}^0$$
For the creation operators the proof consider the dense subspace: $$\mathcal{h}^n_0:=\{\sum_{k=1}^K\lambda_k\alpha_{1,k}\otimes\alpha_{2,k}\otimes\ldots\otimes\alpha_{n,k}:K\in\mathbb{N}\}$$ and reduce quadratic expressions to linear ones via: $$\|\phi\|=\sup_{\|\hat{\eta}\|=1}|\langle \hat{\eta},\phi\rangle|$$
Then an elementary estimate gives: $$\|a(f)\sum_{k=1}^K\lambda_k\alpha_{1,k}\otimes\alpha_{2,k}\otimes\ldots\otimes\alpha_{n,k}\|_{n-1}=\\ =\|\sum_{k=1}^K\lambda_k\sqrt{n}\langle f,\alpha_{1,k}\rangle\alpha_{2,k}\otimes\ldots\otimes\alpha_{n,k}\|_{n-1}\\ =\sqrt{n}\|f\|\cdot\|\sum_{k=1}^K\lambda_k\langle \hat{f},\alpha_{1,k}\rangle\alpha_{2,k}\otimes\ldots\otimes\alpha_{n,k}\|_{n-1}\\ =\sqrt{n}\|f\|\cdot\sup_{\|\hat{\eta}\|=1}|\langle\hat{\eta},\sum_{k=1}^K\lambda_k\langle \hat{f},\alpha_{1,k}\rangle\alpha_{2,k}\otimes\ldots\otimes\alpha_{n,k}\rangle_{n-1}|\\ =\sqrt{n}\|f\|\cdot\sup_{\|\hat{\eta}\|=1}|\langle \hat{f}\otimes\hat{\eta},\sum_{k=1}^K\lambda_k\alpha_{1,k}\otimes\alpha_{2,k}\otimes\ldots\otimes\alpha_{n,k}\rangle_n|\\ \leq\sqrt{n}\|f\|\cdot\sup_{\|\hat{\vartheta}\|=1}|\langle \hat{\vartheta},\sum_{k=1}^K\lambda_k\alpha_{1,k}\otimes\alpha_{2,k}\otimes\ldots\otimes\alpha_{n,k}\rangle_n|\\ =\sqrt{n}\|f\|\cdot\|\sum_{k=1}^K\lambda_k\alpha_{1,k}\otimes\alpha_{2,k}\otimes\ldots\otimes\alpha_{n,k}\|_n$$
Especially the creation operators are bounded. Thus by the bounded linear extension principle: $$\|a(f)\psi_n\|_n=\sqrt{n}\|f\|\cdot\|\psi\|\quad\psi_n\in\mathcal{h}^n$$
Moreover it holds: $$\|a(f)\lambda\Omega\|=\|0\|=\sqrt{0}\|f\|\cdot\|\lambda\Omega\|\quad\lambda\Omega\in\mathcal{h}^0$$