Let $H$ be a Hilbert space, $F$ a Banach space, $G$ a subspace of $H$ and $\varphi: G \to F$ a linear and continuous operator. Show that there is a continuous linear operator $\widetilde\varphi: H \to F$ that extends $\varphi$ and preserves the norm.
My attempt was:
Since $H$ is a Hilbert space, then $H$ is a normed space. Furthermore, since $F$ is a Banach space, $G$ is a subspace of $H$, and $\varphi:G\to F$ is a continuous linear operator, then there is a unique continuous linear extension of $\varphi$ to the close of $G$. Furthermore, the norm of this extension coincides with the norm of $\varphi$. These statements about my hypotheses I have concluded from known results. However, how can I show that there is a linear, continuous operator $\widetilde\varphi: H \to F$ that extends $\varphi$ and preserves the norm?
What is the relation between the closure of $G$ and $H$?
I'll take your question as starting point and assume that $G$ is a closed subspace of $H$. The crucial property here is that closed subspaces of a Hilbert space are complemented: Every element of $H$ can be written uniquely as $x+y$ with $x\in G$ and $y\in G^\perp$. Let $\tilde \varphi(x+y)=\varphi(x)$. This map is clearly linear and $$ \|\tilde\varphi(x+y)\|=\|\varphi(x)\|\leq\|\varphi\|\|x\|\leq\|\varphi\|(\|x\|^2+\|y\|^2)^{1/2}=\|\varphi\|\|x+y\|. $$ Thus $\|\tilde\varphi\|\leq\|\varphi\|$. Since $\tilde\varphi$ coincides with $\varphi$ on $G$, the reverse inequality follows.