Not strongly positive operator

Question

Does anyone know an example of a Hilbert space and a bounded linear operator that is positive definite but its induced bilinear form is not strongly positive?

Thank you very much in advance.

Answer 1

I assume that in this context, $A$ is positive definite if $(x,Ax) \geq 0$ for all $x$ and $Ax = 0 \implies x = 0$.

With this in mind, an example on $\ell^2$: take the operator $$ A(x_1,x_2,x_3,\dots) = \left(\frac{x_1}{1},\frac{x_2}{2},\frac{x_3}{3}, \dots\right). $$ The induced bilinear form is not strongly positive because for $e_n$ (the $n$th standard basis vector) we have $\|e_n\| = 1$ and $(e_n,A e_n) = \frac 1n$.