# Proving that the map between Hilbert spaces is unique.

Theorem: Let $$T:H\rightarrow K$$ be a bounded linear operator between Hilbert spaces. Then there exists a unique bounded linear operator $$T^* :K\rightarrow H$$ such that $$\tag{1} \forall h\in H,k\in K :\langle Th,k\rangle=\langle h,T^*k\rangle.$$

The following proof of existence was given by by lecturer:

Suppose there is another linear operator, $$S :K\rightarrow H$$ such that $$\tag{2} \forall h\in H,k\in K :\langle Th,k\rangle=\langle h,Sk\rangle.$$ Then $$\tag{3} \left\langle h, T^{*} k-S k\right\rangle=\left\langle h, T^{*} k\right\rangle-\langle h, S k\rangle=\langle T h, k\rangle-\langle T h, k\rangle=0 .$$ Thus $$T^{*} k=S k$$ for all $$k\in K$$, and therefore $$T^*=S$$.

What confuses me is that in eq. $$(2)$$ he defined $$S$$ such that $$\langle h,Sk\rangle=\langle h,T^*k\rangle$$ for all $$k\in K$$, and I don't see why eq. $$(3)$$ makes it more obvious that $$T^*=S$$? What exactly is eq. $$(3)$$ showing that is not obvious from eq. $$(1)$$ and $$(2)$$?

• Aug 1, 2022 at 2:37

This is using the fact that if a vector $$v\in H$$ satisfies $$\langle h,v\rangle=0$$ for all $$h\in H$$, then $$v=0$$. (To prove this, just take $$h=v$$ and use the positive-definiteness of the inner product.) Applying this with $$v=T^*k-Sk$$, equation (3) tells you that $$T^*k-Sk=0$$ so $$T^*k=Sk$$.