This might sound oddly phrased. Basically I know that the theorem says that if $T$ is a linear, bounded functional from a Hilbert space $X$ to $\mathbb{C}$ then there is a unique $x_0 \in X$ such that $$T(x) = \langle x,x_0\rangle$$

For the purposes of a question I'm trying to solve about inner products in a space, say $\langle x,z \rangle$, can I simply conclude that there exists a linear functional corresponding to this, say $T_z$, i.e. do the theorem the other way around from a point $z$? Or is this wrong? It seems it is wrong because the theorem seems to be concerned with the other way around. Any clarification would be appreciated - thanks.

  • 2
    $\begingroup$ For any fixed $z$, is $\langle x , z \rangle$ a linear functional in $x$? $\endgroup$ – Theoretical Economist Jan 20 '17 at 17:30
  • $\begingroup$ Hmm, so by the axioms of an inner product it is right? $\endgroup$ – NecroJ Jan 20 '17 at 17:43
  • $\begingroup$ The theorem is stated this way because the other way around it is trivial. $\endgroup$ – Tim B. Jan 20 '17 at 20:04

Let $T_y:X \to \mathbb{R}$ defined by $T(x)=\langle x, y \rangle$ for some $y\in X$.

$T$ is linear, because inner products are linear.

Additionally, $T$ is bounded, because by Cauchy-Schwartz $$\langle x,y\rangle^2 \le \langle x,x\rangle \langle y,y\rangle,$$ which with $K=\sqrt{\langle y,y\rangle}<\infty$ implies $$T_y(x)=\langle x, y\rangle\le K \lVert x \rVert$$ for every $x \in X$.

  • $\begingroup$ Thanks my friend, it has really helped me! $\endgroup$ – NecroJ Jan 20 '17 at 23:24

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