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In physics, the following functional integral is considered \begin{gather} Z[J]= \int Df \exp(-\int d^dx( f\Box f+\lambda f^4 +Jf )) \end{gather} It is usually said that the integration is performed over doubly differentiable functions $f(x)$.

Question 1: Does the result of the integration depend on the function space of integration? E.g., if we choose to integrate over functions that have 3 derivatives, or infinitely many derivatives (in an open set), or holomorphic functions, etc. will the result be different?

There is a rigorous theory of Gaussian measures on Banach spaces.

Question 2: Can the physics functional be considered as a particular case of this Gaussian measure?

The operator $\Box$ is unbounded.

Question 3: Is it essential that there is this quadratic form with an unbounded operator?

One can consider the following modification \begin{gather} Z[A,J]= \int Df \exp(-\int d^dx \int d^d y f(x)A(x,y) f(y)-\int d^dx \lambda f^4 )) \end{gather} where $A$ is an operator in $\mathrm{Aut}(C^L(\mathbb{R}^d))$, where $L$ stands for the smoothness class.

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    $\begingroup$ I don't know about "It is usually said ...". What I hear usually said among physicists is that in the path integral, the set of smooth functions is of measure zero, and the integral is dominated by rough paths. Smooth paths maybe if you're doing a semiclassical approximation. $\endgroup$
    – Michael Engelhardt
    Commented Jan 31 at 3:08
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    $\begingroup$ I think in general properly defining the (functional) measure is hard. What one can do is define it perturbatively around a manageable case, i.e. the Gaussian one. The book by Glimm and Jaffe may be a possible entry (though it's a bit old). $\endgroup$
    – lcv
    Commented Jan 31 at 10:56
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    $\begingroup$ P. Cartier and C. DeWitt-Morette, Functional Integration, Cambridge University Press 2006. $\endgroup$
    – gmvh
    Commented Jan 31 at 10:59

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