Proof. Let \(A=\{x\in H: T(x)=0\}\). Since \(T\) is continuous, \(A\) is a closed subspace of \(H\). If \(A=H\), then \(T(x)=0=(x,0)\) for all \(x\in H\), which proves the assertion.

Now, suppose that \(A\neq H\). By Projection theorem, \(\exists \) a non-zero vector \(z\in A^{\perp }\). Define \[y=\overline {T(z)}||z||^{-2}z\] we show that this \(y\) works. If \(x\in A\), then \(T(x)=0=(x,y)\hspace {.5cm} \text {since}\hspace {0.3cm} y\in A^{\perp }\)
If \(x=\alpha z\), then \begin {align*} T(x) &=T(\alpha z)=\alpha T(z),\hspace {0.3cm} \text {since}\hspace {0.3cm} T\hspace {0.3cm} \text {is linear}\\ &=\Big (\alpha z,\overline {T(z)}||z||^{-2}z\Big )\\ &=(x,y) \end {align*}

i.e \(T(x)=(x,y)\)

Since the functions \(T(x)\) and \((x,y)\) are linear and agree on both \(A\) and \(z\) they must agree on the space spanned by \(A\) and \(z\). But \(A\) and \(z\) span \(H\) since every element \(w\in H\) can be written as \[w=\Bigg (w-\frac {T(w)}{T(z)}z\Bigg )+\frac {T(w)}{T(z)}z\] Thus, \(T(x)=(x,y)\) for all \(x\in H\).

Now suppose there is another vector \(y'\in H\) with \(T(x)=(x,y')\), then \begin {align*} ||y'-y||^2 &=(y'-y,y'-y)\\ &=(y'-y,y')-(y'-y,y)\\ &=T(y'-y)-T(y'-y)\\ &=0 \end {align*}

and so \(y'=y\), proving uniqueness.

Finally, we prove that \(||y||_H=||T||_{H^*}\) Firstly \begin {align*} ||T| &=\sup _{x\in H}\Bigg \{\frac {|T(x)|}{||x||}\Bigg \}=\sup _{x\in H}\Bigg \{\frac {|(x,y)|}{||x||}\Bigg \}\\ &\leq \sup _{x\in H}\Bigg \{\frac {||x||||y||}{||x||}\Bigg \}\\ &=||y||\qquad (1) \end {align*}

Conversely, \begin {align*} ||T|| &=\sup _{x\in H}\Bigg \{\frac {|T(x)|}{||x||}\Bigg \}=\sup _{x\in H}\Bigg |T\Bigg (\frac {x}{||x||}\Bigg )\Bigg |\\ &\geq \Bigg |T\Bigg (\frac {y}{||y||}\Bigg )\Bigg |\\ &=\Bigg (\frac {y}{||y||},y\Bigg )\\ &=||y||\qquad (2) \end {align*}

Combining (1) and (2) proves that \(||y||_H=||T||_{H^*}\). Concludes the proof. □

Proof. Fix \(x\in H\) and consider \[\varphi _x(y)=\overline {B(x,y)} .\] By (ii) this is linear in \(y\), and by (iii) \(|\varphi _x(y)|\leq C||x||\,||y||\), so \(\varphi _x\in H^*\) with \(||\varphi _x||\leq C||x||\).

By the Riesz representation theorem there is a unique vector, call it \(Ax\), with \[\varphi _x(y)=(y,Ax)\quad \text {for all }y, \qquad \text {that is}\qquad B(x,y)=(Ax,y).\]

\(A\) is linear

For scalars \(\alpha ,\beta \) and all \(y\), \[\big (A(\alpha x_1+\beta x_2),y\big )=B(\alpha x_1+\beta x_2,y) =\alpha B(x_1,y)+\beta B(x_2,y)=\big (\alpha Ax_1+\beta Ax_2,\ y\big ),\] using (i). A vector is determined by its inner products against all \(y\), so \(A(\alpha x_1+\beta x_2)=\alpha Ax_1+\beta Ax_2\).

\(A\) is bounded

Taking \(y=Ax\) in \(B(x,y)=(Ax,y)\) gives \(||Ax||^2=B(x,Ax)\leq C||x||\,||Ax||\), so \(||Ax||\leq C||x||\) and \(||A||\leq C\).

Uniqueness

If \((Ax,y)=(A'x,y)\) for all \(x,y\) then \((Ax-A'x,y)=0\) for all \(y\); taking \(y=Ax-A'x\) gives \(Ax=A'x\). □