# Constructing bounded maps on Hilbert spaces

Let $$H_1, H_2$$ be Hilbert spaces and $$(x,y) \in H_1 \times H_2$$ whith $$x \neq 0$$.

1. It exists $$f \in H_1': \lVert f \rVert = 1 \land f(x) = \lVert x \rVert$$
2. Construct $$T: H_1 \rightarrow H_2$$ such that $$Tx = y$$

I am trying to prove this excercise from my Functional analysis book. For the first, I thought that it must be a corolary of Hahn-Banach but since we do not have subspaces I do the following: take $$g_x \in H_1: g_x(z) = \langle z, x \rangle$$ . For all $$z \in H_1: g_x ( z - \frac{g_x(z)}{g_x(x)}x) = 0 \implies H_1 = \Bbb{K}x + \ker g_x$$ . Define $$f: H_1 \rightarrow \Bbb{K}, f(\lambda x + v) = \lambda \lVert x \rVert$$ . It is easy to see that it is linear and that for all $$(\lambda , v) \in (\Bbb{K}\setminus \{0\}) \times \ker g_x : |f(\lambda x + v)|^2 = |\lambda|^2 \cdot \lVert x \rVert^2 = \lVert \lambda x \rVert^2 = \langle \lambda x, \lambda x \rangle = \langle \lambda x + v , \lambda x + v\rangle - \langle v , v \rangle = \lVert \lambda x + v\rVert^2 - \lVert v \rVert^2 \leq \lVert \lambda x + v\rVert^2 \implies |f(\lambda x + v)| \leq \lVert \lambda x + v \rVert \implies \lVert f \rVert \leq 1$$

Trivially we have $$\lVert x \rVert = |f(x)| \leq \lVert f \rVert \cdot \lVert x \rVert \implies 1 \leq \lVert f \rVert$$, so $$\lVert f \rVert = 1$$ .

The problem is that I do not know how to proceed for 2, I have the intuition that might be related to adjoint operators theory but I do not have idea on which theorems use, so any possible help would be appreciated.

Concerning $$1.$$ the linear functional $$f_x(z)=\langle z,{x\over \|x\|}\rangle$$ satisfies the requirements.
Concerning $$2.$$ take $$Tz={1\over \|x\|}f_x(z)y.$$
• You welcome. The conclusion holds for normed spaces $X\times Y.$ This time for a fixed pair $(x,y)$ with $x\neq 0$ by the Hahn-Banach theorem there is $f_x$ satisfying the requirements. Then $T$ is defined in the same way as for Hilbert spaces. Commented May 10 at 19:19