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Existence - If $f=0$ we can just take $u=0$ and thereby have $f(x) =0= \langle x, 0\rangle$ for all $x \in \mathcal{H}$ .
Suppose now $f \neq 0$ , i.e. $Ker f \neq \mathcal{H}$ .
Recall that, since $f$ is continuous, $Ker f$ is a closed subspace of $\mathcal{H}$ (continuity of $f$ implies that $f^{-1}(0)$ is closed in $\mathcal{H}$ ). It then follows from the orthogonal decomposition theorem that
and as $Ker f \neq \mathcal{H}$ we can find $z \in (Ker f)^{\perp}$ such that $\|z\| = 1$ .
It follows easily from the linearity of $f$ that for every $x \in \mathcal{H}$ we have
and since $z \in (Ker f)^{\perp}$ \begin{eqnarray*} \quad\quad\quad\quad\quad\quad 0 & = & \langle f(x)z -f(z)x, z \rangle \\ & = & f(x)\langle z, z\rangle - f(z)\langle x, z\rangle \\ & = & f(x)\|z\|^2 - \langle x, \overline{f(z)} z\rangle \\ & = & f(x) - \langle x, \overline{f(z)} z\rangle \end{eqnarray*} which implies
The theorem then follows by taking $u = \overline{f(z)} z$ .
Uniqueness - Suppose there were $u_1, u_2 \in \mathcal{H}$ such that for every $x \in \mathcal{H}$
Then $\langle x, u_1 -u_2 \rangle = 0$ for every $x \in \mathcal{H}$ . Taking $x = u_1 - u_2$ we obtain $\|u_1-u_2\|^2 = 0$ , which implies $u_1 = u_2$ . $\square$
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