My question is a bit odd, in fact conceptually it is not difficult, only that it operates on objects that are complex (to me).
I would like to check two types of convergence in the product of the spaces where each space consists of iterative composition of a function from the first order with the function from the last order.
In more precise terms: I have two spaces of continuous functions $C(X,C(Y,\mathbb{R}_+))$ and $C(Y,C(X,\mathbb{R}_+))$, where $X$ and $Y$ are compact metric and each $C(.)$ has the topology of uniform convergence, i.e. defined by the metric
$$d(f_1(x),f_2(x))=sup_{x\in X} \mid f_1(x)-f_2(x) \mid$$
From this pair of spaces $C(X,C(Y,\mathbb{R}_+))$ and $C(Y,C(X,\mathbb{R}_+))$ we can obtain the following sequence of spaces.
First take some $f(.)\in C(X,C(Y,\mathbb{R}_+))$ and some $g(.)\in C(Y,C(X,\mathbb{R}_+))$ and define two "composed" functions (it is not function composition in the usual sense, but related):
$$f^{(1)}\equiv g \circ f: X \rightarrow C(C(X,\mathbb{R_+}),\mathbb{R_+})$$
$$g^{(1)}\equiv f \circ g: Y \rightarrow C(C(Y,\mathbb{R_+}),\mathbb{R_+})$$
Interpretation is that $f^{(1)}$ returns for each $x\in X$ a function/map that is itself a mapping from the space of functions to reals.
Next we iterate, namely apply $g(.)$ to $f^{(1)}$ and $f(.)$ to $g^{(1)}$:
$$f^{(2)}\equiv g^{(1)} \circ f: X \rightarrow C(C(C(Y,\mathbb{R_+}),\mathbb{R_+}),\mathbb{R_+})$$
$$g^{(2)}\equiv f^{(1)} \circ g: Y \rightarrow C(C(C(Y,\mathbb{R_+}),\mathbb{R_+}),\mathbb{R_+})$$
We can do again and again such iteration infinitely often, to obtain a sequence of functions (over functions over functions).
My questions are:
How to show for each iteration that the resulting function is continuous? I.e. no matter the order of iteration $n$, each $f^{(n)}:X\rightarrow C(C(C(....),\mathbb{R_+}),\mathbb{R_+})$ is a continuous function (the same for $g^{(n)}$)
Take now $C(X,C(Y,\mathbb{R}_+)) \times C(Y,C(X,\mathbb{R}_+))$ and unfold for each pair $f(.)$ and $g(.)$ the corresponding sequences as above. This let us to produce the sequence of spaces. Do we we have that when $f_k(.) \rightarrow f(.)$ and $g_k(.) \rightarrow g(.)$ as $k\rightarrow \infty$ then $f^{(n)}_k \rightarrow f^{(n)}$ and $g^{(n)}_k \rightarrow g^{(n)} $ uniformly across $n\in \mathbb{N}$?
Thanks a lot for any hints!!!