# On the isomorphism between bounded sesquilinear forms and bounded operators between two Hilbert spaces

Let $H$ and $K$ be two Hilbert spaces. Let $S(H,K)$ be the vector space of bounded sesquilinear forms $u:H\otimes \overline{K}\to\mathbb{C}$, and let $B(H,K)$ be bounded linear operators from $H$ to $K$. Then we have $$B(H,K)\cong S(H,K)\cong B(K,H)$$ $$\phi\mapsto \langle \phi(\cdot),\cdot\rangle = \langle\cdot,\phi^*(\cdot)\rangle \mapsto \phi^*$$ as vector spaces.

If I am correct, all three spaces are actually Banach spaces. Thus I wonder whether the isomorphism involved are isometries. If not, are they are least bounded in both directions?

• Don't you mean "bounded sesquilinear forms on $H\times K\rightarrow\mathbb{C}$"? – Luiz Cordeiro Feb 15 '14 at 21:20
• You're right, but I feel a little uneasy when you say "sesquilinear form" when $u$ is actually linear (if we stick to the definitions)... – Luiz Cordeiro Feb 15 '14 at 21:29
• @LuizCordeiro: a multilinear function is just a linear function from a tensor product, by the definition. – tomasz Feb 15 '14 at 21:30
• @mezhang If I'm correct, you take the conjugate space $\overline{K}$ so that $u$ becomes linear, right? – Luiz Cordeiro Feb 15 '14 at 21:34

Let $\Phi:B(H,K)\rightarrow S(H,K)$ be the morphism you defined. Given $\phi\in B(H,K)$, you can verify that $\Vert\phi\Vert=\sup\left\{|\langle\phi(x),y\rangle|:x\in H, y\in K, \Vert x\Vert\leq 1,\Vert y\Vert\leq 1\right\}=\Vert\langle\phi(\cdot),\cdot\rangle\Vert=\Vert\Phi(\phi)\Vert$, so $\Phi$ is an isometry. Also, it follows (almost directly) from Riesz representation that $\Phi$ is surjective, hence an isomorphism. In the same manner, you prove that the other function is an isometric (conjugate-linear) isomorphism.