Where does it become apparent in real scalar QFT that the field has to be an operator-valued distribution, as opposed to an operator-valued function?

Question

It its very often stated that in QFT, we don't actually deal with operator valued functions (assign a field operator to each point in space time), but instead with operator valued distributions (in general that means that we have to smear over some test region before we can take expectation values of operators).

I'd like to know where this becomes apparant in scalar real QFT (the most simply case one could imagine). I tried to write the the CCR and see if I can say anything interesting about the non-smeared eigenvalues of the field $\Phi$, but I'm stuck.

Answer 1

Since the commutation relations are $$ [\phi(x,t),\pi(y,t)] = \mathrm{i}\delta(x-y)$$ at least one of $\phi$ and $\pi$ must be a distribution, too, since functions are closed under multiplication and subtraction: $\phi\pi - \pi\phi$ when $\pi$ and $\phi$ are both ordinary functions could not result in a $\delta$-distribution on the r.h.s.