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Let $u\in\mathcal{H}_1$ and $v\in\mathcal{H}_2$ and $T\in\mathcal{B}(\mathcal{H}_1,\mathcal{H}_2)$ be a continuous linear operator on hilbert space. This operator is defined by $Tf:=\langle f,u\rangle_{\mathcal{H}_1}v$. I want to find the adjoint $T^*$.

Let $x\in\mathcal{H}_2$, then $\langle Tf,x\rangle_{\mathcal{H}_2}=\langle \langle f,u\rangle_{\mathcal{H}_1}v,x\rangle_{\mathcal{H}_2}=\langle u,f\rangle_{\mathcal{H}_1}\langle x,v\rangle_{\mathcal{H}_2}$

All my attempts to connect the inner products to obtain a $T^*$ with $\langle x,T^*f\rangle_{\mathcal{H}_1}$ failed so far.

I would appreciate a little push into the right direction.

EDIT: All hilbert space mentioned is real.

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    $\begingroup$ If your scalars are the complex numbers, then your last equation is not correct in general. It is also not needed either, even for the real case, to flip those scalar products. $\endgroup$ – user561348 May 26 '18 at 12:25
  • $\begingroup$ @arugula I forgot to add, that only real scalars are allowed. I made an edit. $\endgroup$ – EpsilonDelta May 26 '18 at 12:27
  • $\begingroup$ Regardless, it is not needed even for the real case. You did it because you were looking for the wrong formula for the definition of the adjoint. $(Tf,x)=((f,u)v,x)=(f,u)(v,x)=(f,\overline{(v,x)}v)$. Therefore, the adjoint is $T^*x=\overline{(v,x)}v$. $\endgroup$ – user561348 May 26 '18 at 12:28
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You are looking for an operator $T^*$ satisfying $\langle f, T^* x\rangle_{\mathcal{H}_1}=\langle Tf, x\rangle_{\mathcal{H}_2}$

As you wrote (you basically solved it),

$$\langle Tf,x\rangle_{\mathcal{H}_2}=\langle \langle f,u\rangle_{\mathcal{H}_1}v,x\rangle_{\mathcal{H}_2}=\langle u,f\rangle_{\mathcal{H}_1}\langle x,v\rangle_{\mathcal{H}_2}=\langle \langle x,v\rangle_{\mathcal{H}_2}u,f\rangle_{\mathcal{H}_1}$$

So $T^*x = \langle x,v\rangle_{\mathcal{H}_2}u$

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