[Math] Why is the delta function the continuous generalization of the kronecker delta and not the identity function

Question

In a discrete $n$ dimensional vector space the Kronecker delta $\delta_{ij}$ is basically the $n \times n$ identity matrix. When generalizing from a discrete $n$ dimensional vector space to an infinite dimensional space of functions $f$ it seems natural to assume that the generalization of the Kronecker delta should be an identity operator
$$
\operatorname{I} f = f
$$
However it is said that the continuous generalization of the Kronecker delta is the Dirac delta function
$$
\int_{-\infty}^\infty \delta(x – y) f(x)\, dx = f(y)
$$
Why is that the case? What is "wrong" with the simple identity operator?

UPDATE: What I mean with "simple" identity operator is: Why is the identity operator not simply the scalar number "1", but the delta function instead?

UPDATE2: To make it more clear: Why is the continuous generalization of the Kronecker delta
$$
\int_{-\infty}^\infty \delta(x – y) f(x)\, dx = f(y)
$$
and not
$$
1 \cdot f(x) = f(x)
$$
?

Answer 1

The identity operator is the same in both cases.

For the discrete case $$ \sum_{j=1}^n \delta_{ij} x_j = x_i $$ and in terms of operators, $I(\mathbf{x}) = \mathbf x$ .

For the continuous case $$ \int_{-\infty}^\infty \delta(x - y) f(x)\, dx = f(y) $$ and in terms of operators $I(f) = f$.