Reproducing Kernel Hilbert Space, Condition (i). Barry Simon problem 4.

Question

I am reading Barry Simon's Analysis. In it he defines: A reproducing kernel Hilbert space to be a Hilbert space , $\mathcal{H}$, of functions , $f$ on a set, $E$, so that

i) For any $f$, there is $x\in E$ with $f(x)\not=0$

ii) For any $x\in E$, there is $f\in \mathcal{H}$ so $f(x)\not=0$

iii) For any $x,y\in E$, there is $f$ in $\mathcal{H}$ so $f(x)\not=f(y)$

iv) For any $x\in E$, there is $C_x$ so that $|f(x)|\leq C_x|f|_\mathcal{H}$

Just a couple (simple) questions:

• Why does Simon phrase condition i) the way he does?
  (Isn't this first condition just saying $\mathcal{H}$ is not the trivial Hilbert space, the Hilbert space consisting of just the zero function $\mathbf{0}:E \mapsto 0$?)
• Since the zero function $\mathbf{0}$ maps every point in $E$ to $0$ is $\mathbf{0}$ not in the reproducing kernel Hilbert space?

Answer 1

You are probably imagining $\mathcal H$ to be the dual space of $\mathcal H$. It is just some family of functions on $\mathcal H$ satisfying the four properties. If $\mathcal H=\{1\}$ the i) will fail whether or not $\mathcal H=\{0\}$.

Yes, $0$ is not in the reproducing kernel Hilbert space.

Answer 2

After thinking about this some more, I think in condition i) Simon means "for any nonzero function in the Hilbert space." I am guessing he states conditions i) and ii) the way he does to show the symmetry of these first two conditions. If correct, then perhaps a better way to state them might be:

i) For any $\mathbf{}f\in \mathcal{H}\setminus\{\mathbf{0}\}$, there is $x\in E$ with $f(x)\not=0$

ii) For any $x\in E$, there is $f\in \mathcal{H}\setminus\{\mathbf{0}\}$ so $f(x)\not=0$

Deleting the zero function from the Hilbert space is clearly redundant in ii), however it seems necessary to make i) technically accurate.