A Lagrangian density, a polynomial in a classical field and its derivatives, can be used to specify a dynamics of a conservative classical field, without too much restriction on differentiability of the initial classical field configuration.

A quantum field, however, is a higher-order mathematical structure —an initial quantum field state specifies an infinite system of incompatible probability measures *over* classical field configurations— so why do we not need a higher-order object than a Lagrangian density to specify the dynamics?

If we work with a crude regularization, a cutoff \(\Lambda\), for example, effectively using the quantum field in convolution with a smooth cutoff function \(f_\Lambda(x)\), \(\hat\phi_\Lambda(x)=[\hat\phi\star f_\Lambda](x)\), the regularized quartic Lagrangian density would be \(:\!\frac{\lambda}{4!}(\hat\phi^\dagger_\Lambda(x)\hat\phi_\Lambda(x))^2\!:\), but we could also use a quartic Lagrangian density \(:\!\sum_i\frac{\lambda_i}{4!}(\hat\phi^\dagger_{\Lambda_i}(x)\hat\phi_{\Lambda_i}(x))^2\!:\), or a continuum version, \(:\!\int\frac{\lambda(z)}{4!}(\hat\phi^\dagger_{\Lambda(z)}(x)\hat\phi_{\Lambda(z)}(x))^2\mathrm{d}z\!:\). Is there a proof somewhere that these and all comparable constructions will be no different from the single \(\Lambda\) case, no matter how we take the limits \(\Lambda_i,\Lambda(z)\rightarrow \infty\) and no matter how we take the coupling, mass and field renormalizations? *ADDED: From an effective field theory perspective the finite cutoff case is of interest in itself, for which case the multiple cutoff field examples are distinct from the single cutoff field example.* [Preferably take this last paragraph to be illustrative, not as part of the question.]