The issue is one needs a Banach space for the Fr ́echet derivative. The C∞(Rn) is not a Banach space so the Frechet derivative cannot be defined directly.

C∞(Rn) is a Frechet space and it can be viewed as the projective limit of spaces on a sequence of expanding compact sets whose union covers R^n the smooth functions in each of these spaces is a Banach space. That with the norm on each of the bing the C^infinity sup norm should be sufficient for your needs.

http://www-users.math.umn.edu/~garrett/m/fun/notes_2012-13/02_spaces_fcns.pdf

C^infinity(R^n) is not a normed space. It's topology is defined by a family of semi-norms, a Frechet space.The sup( ) is not a norm on the smooth functions on a non compact set without other conditions on the function, line the function and all of its derivatives converging to 0. There are smooth functions that coverage uniformly in the sup norm which but the limit is not smooth.

http://www.math.ucla.edu/~tao/resource/general/131bh.1.03s/week45.pdf

The necessary topology is the one that is the sup over all derivatives and over the space on which the functions are defined. That topology makes C^infinity into a Frechet space.

But for your question it doesn't matter since it is a local question and it can be answered in considering a compact subset containing the pointing its interior of interest and which will be a normed (Banach ) space where the topology of uniform convergence of the functions all all its derivatives on the compact subset. In those spaces the Frechet derivative is well defined and that should be sufficient to answer your question.

See for example,

https://www.amazon.com/Distributions-Operators-Graduate-Texts-Mathematics/dp/0387848940/ref=sr_1_8?s=books&ie=UTF8&qid=1548986826&sr=1-8&keywords=distribution+mathematics

https://www.amazon.com/Distributions-Differential-Equations-Harmonic-Universitext-ebook/dp/B00FB25Q5G/ref=sr_1_3?s=books&ie=UTF8&qid=1548987075&sr=1-3

.

To fix and simplify the ideas, let n=1, Y=[0,1]. The space C_P(^(infinity)(Y) is a normed vector space with respect to the sup-norm, since Y is compact (the sup over R coincide with the sup over Y, for any periodic (continuous) function (of period 1) in the latter space). Thus, it seems that the problem is correctly formulated. An idea for approaching it might be as follows. Without loss of generality, we can assume that Y =[0,2*pi] or [-pi,pi]. Then for any x in Omega, one can write:

psi(x)(y)=a_0(x)+Sum(n=1 to infinity)(a_n(x)*cos(ny)+b_n(x)* sin(ny)), y in Y or in R, where a_n(x), b_n(x) are the Fourier coefficients, given by the corresponding integrals on [-pi, pi]. For any fixed x in K:=supp(psi), the trigonometric series written above converges uniformly and absolutely on R (with respect to variable y), since psi(x) is periodic and of class C^(1) (cf Dirichlet's theorem). It is obvious that any partial sum of the above function series is of class C^(infinity) on R^2. On the other hand, observe that on the space C_P(^(infinity)(Y) one can define the norms

||f||_k:=sup|f^(l)(y)|, (k in N), the sup being taken with respect to y in Y (or in R) and to l in {0,1,...,k}. These norms define the topology of uniform convergence together with all derivatives in the space C_P(^(infinity)(Y). It seems that the latter space is a Frechet space, endowed with this family of norms. These remarks do not solve the problem, but might contain some information. The problem is that we cannot derivate term by term with respect to y the trigonometric series written above (by derivation with respect to y, the convergence of the obtained series could be lost). A possible further approach is to consider weak derivatives of psi=

psi(x)(y)=psi(x,y), in the sense of (regular) distributions theory.

Yes one can define a norm on C^infinity by uniform convergence - sup norm. However, C^infinity is not complete in this norm. That is there are smooth functions that converge uniformly to a function and the limit is not smooth. In general lack of completeness is an issue. Defining the metric by semi norms of uniform convergence of all deviates is a complete gives a complete topology on the smooth functions.

http://www.math.ucla.edu/~tao/resource/general/131bh.1.03s/week45.pdf

However, the second space which has no topology defined is periodic smooth functions on the unit cube. This can be identified in a natural with with the smooth functions on a Torus which is compact. The norm that defines uniform convergence is again not sufficient for the space to be complete. If completeness is important one has look at uniform convergence of all the derivatives in which case the space becomes the projective limit of Banach spaces and the metric topology is complete. The space in the original question

the smooth functions on the cube are the test functions for distribution theory over the Torus. The topology is defined by the semi-norms of uniform convergence of all derivatives which makes it a complete metric space.

Similar results hold that hold on R^N when Fourier transform is replaced by the Fourier transform over the Torus (since the Torus is a group) which is the Fourier series over the Torus. This allow the definition of periodic distributions which have similar properties as distributions on R^N but depend on the fact that the dual group of an N-dimensional Torus is the N fold product of the integers.

Here is a discussion for one dimensional (circle). However, the same results hold for the Torus,

http://users.jyu.fi/~salomi/lecturenotes/FA_distributions.pdf

Octav Olteanu Truman Prevatt Thanks very much for the responses. I will think about it.