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This question concerns the theorem that says for any bounded sesquilinear form on a hilbert space $H$, $u:H\times H \to \mathbb{F}$ we can uniquely associate it with bounded operators $A,B$ such that $u(h,k) = \langle Ah,k\rangle = \langle h,Bk\rangle$ for all $h,k\in H$.

The proofs I've seen for this proceed the same way in defining a functional $L_h: H \to \mathbb{K}$ by $L_h(k) = \overline{u(h,k)}$, applying the Riesz representation theorem to get $f\in H$ such that $\langle k,f\rangle = L_h(k) = \overline{u(h,k)}$, then defining $A$ to be the operator satisfying $Ah=f$ whence $\langle Ah,k \rangle = \overline{\langle k,Ah\rangle} = \overline{\langle k,f \rangle} = u(h,k)$.

What I'm confused about is how this uniquely associates a single $A$ to $u$ rather than associating a different $A$ to every pair of $h$ and $k$, since the Riesz representation theorem would give a different $f$ for different $u$ and for different $h$ different $A$ would make $Ah=f$.

summersfreezing
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  • For each $h\in H$ you find $f\in H$ and define $Ah = f$, which defines an operator $A:H\rightarrow H$, which by uniqueness of $f$ is unique again. – stange Jan 04 '24 at 13:41
  • Suppose $\alpha\in\Bbb{C}$, $h_1, h_2\in H$ and you found $f_1, f_2$ such that $\langle f_1, k\rangle = u(h_1, k), \langle f_2, k\rangle = u(h_2, k)$ for all $k$. Then $\langle \alpha f_1 + f_2, k\rangle = \alpha\langle f_1,k\rangle+\langle f_2,k\rangle= \alpha u(h_1, k) + u(h_2, k)=u(\alpha h_1 + h_2, k)$ for all $k$. This shows $A(\alpha h_1 + h_2)$ must be $\alpha f_1 + f_2=\alpha Ah_1 + Ah_2$ by the uniqueness from the Riesz representation theorem. So $A$ is linear. – Chad K Jan 04 '24 at 14:30

1 Answers1

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The $f$ you find in your argument depends on $h$ and is unique for a given $h$ (and all $k$), hence you can write it as $Ah$. That gives you $u(h,k)=\langle Ah,k\rangle$ for all $h,k$.

The uniqueness also implies that $A$ is linear because $$\langle aAh_1+Ah_2,k\rangle=a\langle Ah_1,k\rangle+\langle Ah_2,k\rangle =au(h_1,k)+u(h_2,k)=u(ah_1+h_2,k). $$ As this holds for all $k$, the uniqueness in the Riesz representation gives you that $A(ah_1+h_2)=aAh_1+Ah_2$. You also get that $A$ is bounded, since $$ |\langle Ah,k\rangle|=|u(h,k)|\leq\|u\|\,\|h\|,\|k\|. $$ As $\|A\|=\sup\{|\langle Ah,k\rangle|:\ \|h\|=\|k\|=1\}$, you get that $\|A\|\leq\|u\|$.

Martin Argerami
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