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As a practice exercises(not an assignment question) for one of the papers I am doing currently at university we are asked to show the following;

I $T:H \rightarrow K$ is a bounded linear operator from between two Hilbert spaces. Show that there exists a unique bounded linear operator $T^*:K\rightarrow H$ such that

$$ \langle Th,k\rangle = \langle h,T^*k \rangle\ \ \ \ \forall h\in H,\ \ \forall k\in K. $$

Uniqueness is easy, for if there existed $S,P \in \mathcal{L}(K,H)$ satisfying this property then we would have

$$ \langle h, Pk-Sk\rangle =0 \ \ \ \ \forall h \in H \ \ \ k\in K $$ in particular for $Pk -Sk \in H$ we would have

$$ \langle Pk-Sk, Pk-Sk\rangle = 0 \ \ \ \forall k\in K $$ which means $Pk= Sk$ for all $k\in K$ so they are the same.

Proof of existence

We start by fixing $k \in K$. Now we define the linear functional $L_k : H \rightarrow \mathbb{F}$ by

$$ L_k(h) := \langle Th,k \rangle, \ \ \ \ h \in H $$ ($L_k$ is linear by linearity of $T$ and linearity of inner products and bounded by Cauchy-Schwarz.) Now by the Riesz - Representation Theorem we know there exists a unique $v \in H$ such that

$$ L_k(h) = \langle h,v \rangle , \ \ \ \ \ \forall h \in H. $$

Notice that the $k \in K$ that we fixed was arbitrary so for each $k \in K$ there exists a unique $v_k \in H$ such that

$$ \langle Th,k\rangle = L_k(h) = \langle h,v_k \rangle , \ \ \ \ \ \forall h \in H. $$

Now we define the function $T^* :K \rightarrow H$ where

$$ T^* (k) := v_k, \ \ \ \ k \in K. $$

Now we claim that $T^*$ is linear and bounded.

We start by showing that $T^*$ is linear. So we fix $k,g \in K$ and $\lambda \in \mathbb{F}$, the case where $k = g = 0$ is trivial so we assume $k,g \neq 0$, then we need to show that

$$ T^*(\lambda k +g) = \lambda T^*k +T^*g. $$

Or by the uniqueness of $T^*(k)$ for all $k \in H$ it is enough to show show that

$$ \langle Th,\lambda k +g \rangle = \langle h, \lambda T^*k +T^*g \rangle, \ \ \ \ \forall h \in H $$ Now for arbitrary $h \in H$ we have

$$ \langle Th, \lambda k +g \rangle = \bar{\lambda}\langle Th,k \rangle + \langle Th,g \rangle = \langle h, \lambda T^*k\rangle +\langle h, T^*g \rangle = \langle h, \lambda T^*k +T^*g \rangle $$

as needed for linearity.

Now for boundedness we notice for $k \in K$, (the case where $k = 0$ is trivial so we may assume $ k \neq 0$)

$$ ||T^*k||^2 = \langle T^*k,T^*k \rangle = \langle k , T(T^*k) \rangle \leq ||k||\ ||T(T^*k) || \leq ||k||\ ||T||_{\text{op}}\ || T^*k || $$

By simple rearranging it follows that

$$ ||T^*k|| \leq ||T||_{\text{op}}\ ||k||. $$

As a bonus we also notice that the from infemum characterization of the operator it follows that

$$ ||T^*||_{\text{op}}\leq ||T||_{\text{op}}. $$

QED

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    $\begingroup$ Do you know the Riesz representation theorem for Hilbert spaces? Fix $x \in H$. Use $x$ and $T$ to find a linear functional on $K$. Now note that every linear functional is of a certain kind by the Riesz representation theorem : it is given by the inner product with a certain element of $K$. The map $T^*$ should take $x$ to this element. Now show that $T^*$ is linear and bounded. (If further hint is required, I will write an answer.) $\endgroup$ Jun 6, 2018 at 2:48
  • $\begingroup$ Yes, I have encountered the Riesz Representation theorem, that was another exercise. I did not know to look in its direction for this problem though. Thanks very much. I will use it to prove this then add my proof and edit my question to a proof check instead. $\endgroup$ Jun 6, 2018 at 2:55
  • $\begingroup$ You are welcome! Please return for doubts. $\endgroup$ Jun 6, 2018 at 2:56
  • $\begingroup$ I've added my proof, Would you be so kind as to check it? $\endgroup$ Jun 6, 2018 at 5:14
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    $\begingroup$ It is correct. Well done, $+1$. $\endgroup$ Jun 6, 2018 at 5:17

1 Answer 1

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The accepted proof is the proof presented above.

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