An operator on the space of compactly supported sequence does not satisfy a given property.
I assume the confusion around notation is already resolved, so I'll focus on the representability of $\lambda$ as the inner product against a fixed square-summable sequence $f$.
Suppose there were such $f \in \ell^2$, then, for each basis vector/sequence $e_i$, we would have to have $\lambda(e_i) = 1 = \sum_{j=1}^\infty e_i(j)\overline{f(j)} = \overline{f(i)}$. Thus, $f = (1,1,...)$ but such sequence isn't in $\ell^2$. Therefore, no such $f$ exists.