In a separable Hilbert space $\mathcal{H}$, given a complete orthonormal basis $\{|e_i\rangle\}$, the identity operator can be written as $\mathbb{1} = \sum_i |e_i\rangle\langle e_i|$. Now if this Hilbert space is reproducing, can I write the identity operator as $\mathbb{1} = \sum_i |k_{x_{i}}(\cdot)\rangle\langle k_{x_{i}}(\cdot)|$ (or an integral as $\int dx |k_{x}(\cdot)\rangle\langle k_{x}(\cdot)|$), where $|k_{x_{i}}(\cdot)\rangle=|k(x_{i},\cdot)\rangle$ is the reproducing kernel function? (I know they are not orthonormal since the kernal gramian matrix $k(x_i,x_j)\neq\delta_{ij}$ in general, but I just want to know whether I can still plug in $\sum_i |k_{x_{i}}(\cdot)\rangle\langle k_{x_{i}}(\cdot)|$ at anywhere in an equation, just as I can do with $\sum_i |e_i\rangle\langle e_i|$.)

If not, why the following proof is wrong?

$\forall |\psi\rangle \in \mathcal{H}$, $|\psi\rangle = \sum_i |e_i\rangle\langle e_i|\psi\rangle$, now the inner product $\langle e_i|\psi\rangle$ may be expressed as $\int e_i^*(x)\psi(x)dx$ (or a summation $\sum_j e_i^*(x_j)\psi(x_j)$), but since $k(x_{i},\cdot)$ is reproducing, we should have $\langle e_i|\psi\rangle=\int dx\langle e_i|k_x\rangle\langle k_x|\psi\rangle$, so $|\psi\rangle = \sum_i |e_i\rangle\langle e_i|\psi\rangle=\int dx\sum_i |e_i\rangle\langle e_i|k_x\rangle\langle k_x|\psi\rangle$, now since $\sum_i |e_i\rangle\langle e_i| =\mathbb{1}$, we have $|\psi\rangle=\int dx|k_x\rangle\langle k_x|\psi\rangle$, which is valid for all $|\psi\rangle \in \mathcal{H}$, hence $\int dx|k_x\rangle\langle k_x|=\mathbb{1}$.

If yes, there seems to be a strange contradiction:

Consider the kernel Gramian matrix's elements $G_{ij}=k(x_i,x_j)=\langle k_{x_{i}}(\cdot)|k_{x_{j}}(\cdot)\rangle$, now if I plugin $\mathbb{1} = \sum_m |k_{x_{m}}(\cdot)\rangle\langle k_{x_{m}}(\cdot)|$ between the inner product, I will have $G_{ij}=\sum_m \langle k_{x_{i}}(\cdot)|k_{x_{m}}(\cdot)\rangle\langle k_{x_{m}}(\cdot)|k_{x_{j}}(\cdot)\rangle=\sum_m k(x_i,x_m) k(x_m,x_j)=G_{im}G_{mj}$, so that means $G=G^2$ or $G=G^n, \forall n$, which means that all eigenvalues $\lambda_i$ of $G$ must satisfy $\lambda_i^n=\lambda, \forall n$, so all eigenvalues of kernel Gramian matrix $G$ must be $1$ or $0$, which is apparently wrong.

So I must be wrong somewhere.

Thanks in advance!

  • $\begingroup$ It's very difficult to follow your notations. If I understand correctly, in the first sentence of your proof you assume that the inner product is always an integral. If this is the case, then this is a wrong assumption, I think. $\endgroup$ – erz May 6 '17 at 22:13
  • $\begingroup$ If I understand correctly, the part that starts with "we should have" is wrong (which is, by the way, kind of expected). $\endgroup$ – Mateusz Kwaśnicki May 6 '17 at 22:49

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