# Is there a deeper connection between the two Riesz's Representation Theorems?

I have been reading Kreyszig's Functional Analysis when I encountered two versions of Riesz's Representation Theorems:

(1) Every bounded linear functional $f$ on a Hilbert space $H$ can be represented in terms of the inner product, namely $f(x)=\langle x,z\rangle$, where $z$ is uniquely determined by $f$ and $\|z\|=\|f\|$.

(2) Let $H_{1},H_{2}$ be Hilbert spaces and $h:H_{1}\times{}H_{2}\rightarrow{}K$ a bounded sesquilinear form. Then $h$ has a representation $h(x,y)= \langle Sx,y \rangle$, where $S:H_{1}\rightarrow{}H_{2}$ is a bounded linear operator. $S$ is uniquely determined by $h$ and has norm $\|S\|=\|h\|$.

The book used (1) to deduce (2), but I am curious whether you can deduce (1) given (2)? Is there a way to understand (1) as a special case of (2)?

Also, I find these two theorems very similar to each other. So I'm curious if is there a deeper connection between them, other than the superficial similarity in their names and their mutual concern for norm equality?

Sorry if this is a vague questions...

Thank you!!

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Try to do the real case first. Simply consider $h: H \times H^\ast \to \mathbb{R}$ given by $h(x,f) = f(x)$. After you got that one right, you need to think about where to complex conjugate. –  t.b. May 5 '12 at 11:14
@t.b.: it looks to me that you are using that $H^*$ is a Hilbert space, which usually requires 1). –  Martin Argerami May 5 '12 at 14:13
@Martin: you're right that there's a word to say here. It suffices to check the parallelogram law for the norm on $H^\ast$. To this end, you can assume that the functionals $f$ and $g$ are linearly independent and observe that their joint kernel $K$ then has codimension $2$ and yields an orthogonal decomposition $H = K \oplus \mathbb{C}^2$. This essentially reduces the problem to $\mathbb{C}^2$ where linear algebra does the rest. Of course, this consideration also yields (1) quite easily but you don't need to use it formally. –  t.b. May 5 '12 at 15:31
@t.b. Thank you for your reminder! I'm guessing in the complex case you mean $h(ax,bf):=af(b^{\star}x)=<Sx,f>=<x,S^{\star}f>=<x,z>$, where $h$ is a sesquilinear form, am I right on this? –  Vokram May 6 '12 at 3:38
@Vokram: sorry for replying so late, not quite. I'll elaborate into an answer tomorrow if there's not an answer already. –  t.b. May 6 '12 at 22:07