Riesz Representation Theorem states: For each $f \in \mathcal H^*$ there exists a unique $y \in \mathcal H$ such that $f = f_y$, where $f_y: \mathcal H \to \mathbb K, f_y(x) := \langle x \,,\,y\rangle$.

I saw the following proof:

Let $y \in \mathcal H^*$, $f \neq 0$ be arbitrary. Denote by $N$ the kernel ker$(f)$ of $f$ and let $z' \in \mathcal H$ with $f(z') = 1$. Since $N$ is closed and $ \mathcal H$ is complete, we can decompose $z'$ as $u + z$ with $u ∈ N$ and $z ∈ N^{⊥}$. Then $1 = f(z') = f(u) + f(z) = f(z)$. For arbitrary $x \in \mathcal H$, we have $x - f(x)z \in$ ker$(f)$, which implies $f(x) = \langle f(x)z \,,\,||z||^{-2}z\rangle = \langle (x-f(x)z) + f(x)z\,,\,||z||^{-2}z\rangle = \langle x \,,\,||z||^{-2}z\rangle = f_y(x)$, for $y:= ||z||^{-2}z$. This shows that claim.

I don't understand the second equality, because this does already imply that $\langle (x-f(x)z)\,,\,||z||^{-2}z\rangle = 0$ which is equivalent to $\langle x \,,\,||z||^{-2}z\rangle = f(x)$, which is was still to show.

  • $\begingroup$ What you wrote doesn't make sense, you have the scalar $f(x)$ being added to the vector $x - f(x)z$ $\endgroup$
    – D_S
    Aug 10 '17 at 18:33
  • 1
    $\begingroup$ There's a typo, it ought to be $$\dotsc = \langle (x - f(x)z) + f(x)\underbrace{\cdot z}_{\uparrow}\;, \lVert z\rVert^{-2} z\rangle = \dotsc$$ $\endgroup$ Aug 10 '17 at 18:52
  • $\begingroup$ To show that $x-f(x)z$ is in the kernel of $f$, use that $f$ is linear and sends $z$ to $1$, to compute $f(x-f(x)z)=f(x)-f(x)f(z)=f(x)-f(x)=0$. $\endgroup$ Aug 10 '17 at 20:35

I kinda overlooked something: Notice that $x - f(x)z \in$ ker$(f) = N$, i.e. $x - f(x)z ⊥ z$ as $z ∈ N^{⊥}$. But from this we can already follow that $\langle (x-f(x)z)\,,\,||z||^{-2}z\rangle = ||z||^{-2}\langle (x-f(x)z)\,,\,z\rangle = 0.$


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