I am trying to prove the existence of adjoints of bounded linear operators on Hilbert spaces:

If $H,H'$ are Hilbert spaces and $u \in B(H,H')$ then there exists a unique $u^\ast \in B(H',H)$ such that for all $h \in H, h' \in H'$:

$$ \langle u(h), h'\rangle = \langle h, u^\ast (h') \rangle$$

Here is what I have so far:

Since $u$ is bounded and linear, $h \mapsto \langle u(h), h'\rangle$ is in $H^\ast$. Therefore by the Riesz representation theorem there exists $h_0 \in H$ such that

$$ \langle u(h), h'\rangle = \langle h, h_0\rangle$$

We define

$$ u^\ast (h') = h_0$$

and claim that $u^\ast$ is linear and continuous.

Linearity: Consider $u^\ast (\lambda h' + \mu h'')$. By how $u^\ast$ was defined, for all $h \in H$:

$$ \langle h, u^\ast (\lambda h' + \mu h'') \rangle = \langle u(h), \lambda h' + \mu h''\rangle$$

Now I get stuck because the right argument is conjugate linear, not linear.

How can I show that $u^\ast$ is linear?

  • $\begingroup$ The conjugation is reversed, since you'll do the swapping twice. $\endgroup$
    – Pedro
    Dec 8, 2014 at 1:41

1 Answer 1


You are doing fine: $$ \langle h, u^\ast (\lambda h' + \mu h'') \rangle = \langle u(h), \lambda h' + \mu h''\rangle =\bar\lambda\langle u(h),h'\rangle+\bar\mu\langle u(h),h''\rangle\\ =\bar\lambda\langle h,u^*(h)'\rangle+\bar\mu\langle h,u^*(h'')\rangle=\langle h,\lambda u^*(h')+\mu u^*(h'')\rangle. $$

  • $\begingroup$ Thank you. Why didn't I see that. Now I'm stuck on how to finish the proof. I want to do something like: $$ \\$$ Since for all $h \in H$: $$ \langle h, u^\ast (\lambda h' + \mu h'' )\rangle = \langle h, \lambda u^\ast (h') + \mu u^\ast (h'')\rangle$$ it holds in particular for $h = u^\ast (\lambda h' + \mu h'' )$. Hence $$ \|u^\ast (\lambda h' + \mu h'' )\|^2 = \|\lambda u^\ast (h') + \mu u^\ast ( h'' )\|^2$$ which implies $u^\ast (\lambda h' + \mu h'' ) = \lambda u^\ast (h') + \mu u^\ast ( h'' )$ but this last step is clearly wrong. $\endgroup$
    – user167889
    Dec 8, 2014 at 1:57

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