I have defined the map $\alpha:H\to H^*$ given by $$(\alpha x)(y) = \langle y, x\rangle.$$ I want to use it to show that:

(1) it is an isometric imbedding of $H$ onto $H^*$,

(2) and that $\alpha(\lambda x + \mu z) =\bar{\lambda} \alpha x + \bar{\mu}\alpha z$.

Where $H$ is a Hilbert space and $H^*$ is its dual space.

I already that for every bounded linear functional $x^*$ on a Hilbert space $H$ there exists a unique element $z$ of $H$ such that $x^*(x) = \langle x,z\rangle$ for all $x\in H$


1 Answer 1


(1) You need to show that the operator norm of $\alpha$ is equal to the norm of $x$ in $H$, this proves isometry. The property that $H$ is isometrically isomorphic to $H^*$ is the famous Riesz Representation Theorem.

We have $$\|\alpha x\| = \sup_{\|z\| = 1}|(\alpha x)(z)| = \sup_{\|z\| = 1}|\langle z,x\rangle|\le \|x\|.$$ Equality is indeed reached, by choosing $z = x/\|x\|$. So $\|\alpha x\| = \|x\|$, proving isometry.

(2) Follows directly from the properties of the inner product: Let $z\in H$ arbitrary: $$\alpha(\lambda x + \mu y)(z) = \langle z, \lambda x + \mu y\rangle = \langle z, \lambda x\rangle + \langle z,\mu y\rangle =\bar \lambda \langle z, x\rangle + \bar\mu\langle z,y\rangle =\bar \lambda \alpha(x)(z) + \bar\mu\alpha(y)(z), $$ using the linearity and conjugate symmetry of the inner product.


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