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# Bases for Spaces of Smooth Functions

The papershows that for a compact manifold \$M\$, the Frechet space \$C^infty(M)\$ is always linearly isomorphic to the space \$mathcal s\$ of rapidly deceasing sequences.If you consider the \$L^2\$ orthonormal basis \$phi_k\$ for the Laplacian of a Riemannian metric on \$M\$ as suggested by Liviu Nicolaescu, then any \$fin C^infty(M)\$ is of the form \$f=sum_k f_kphi_k\$, and \$f_kin ell^2\$ initially. But \$1Delta\$ (geometric \$Delta\$ here) is an isomorphism between the Sobolev spaces \$H^k(M)\$ and \$H^k-2(M)\$ for each \$k\$. By Weyl's formula the eigenvalues \$lambda_k\$ of \$Delta\$ satisfy \$lambda_k sim C k^2/dim(M)\$ for \$kto infty\$; see page 155 of the book of Chavel, Eigenvalues in Riemannian geometry'. Since \$bigcap_k H^k(M)=C^infty(M)\$ on a compact manifold, we see that \$\$(1Delta)^m f = sum_k f_k(1 lambda_k)^mphi_k\$\$ with coefficients again in \$ell^2\$, for each \$m\$. Thus the coefficients \$f_k(1Ck^2/dim(M))^min ell^2\$ for each \$m\$, and the \$f_k\$ are rapidly decreasing. Moreover, any rapidly decreasing sequence of coefficients gives a function in \$C^infty(M)\$. This proves again that \$C^infty(M)cong mathcal s\$, even with the basis of eigenfunctions for any Laplacian.

1. The number of functions \$f: cal P_n to 1, 2, dots, m\$ such that \$f(A cap B) = minf(A), f(B)\$ (Putnam 1993)

For notational ease, let \$N = 1,ldots,n\$, \$N_i = [n]setminusi\$, and \$M = 1,ldots,m\$.First, assume that \$f : mathcalP_n to M\$ and that \$f(A cap B) = minf(A),f(B)\$.Choose \$i in N\$ and \$j in M\$. It is certain that \$f(N) in M\$, so suppose \$f(N) = j\$. Then \$f(N cap N_i) = f(N_i) = minf(N),f(N_i)\$. It follows that for all \$i\$, \$f(N_i) le f(N) = j\$.As an intermediate step, I will prove via induction that when we specify \$f(N) = j\$ for some \$j in M\$ and \$f(N_i)\$ for all \$i in N\$, \$f\$ has been fully specified and is well-defined. Assume that \$f(S) = min_i

otin Sf(N_i)\$ for subsets \$S subset N\$ where \$m le |S| le n-1\$. Now, consider subsets \$S in N\$ where \$|S| = m-1\$. Let \$A\$ and \$B\$ be subsets of \$N\$ such that \$|A|,|B| ge m\$ and \$A cap B = S\$.Note that the condition \$f(A cap B) = minf(A),f(B)\$ is enough to guarantee that the image of \$S\$ under \$f\$ is defined at all, because we can always find at least two sets \$A,B\$ with the above requirements such that \$A cap B = S\$. For instance, take \$a,b in N\$ such that \$a,b

otin S, a

e b\$; then let \$A = S cup a\$ and \$B = S cup B\$. Then \$A

e B\$ and \$|A|,|B| ge m\$ and \$A cap B = S\$, and since \$|S|

2. Arity of Primitive Recursive Functions

You can indeed define \$g(n)=h(n,f(n))\$ (as I assume you intended to write) -- but in order to argue that this \$g\$ is primitive recursive, you need to already know that \$f\$ (as well as \$h\$) is primitive recursive, and for that you need to apply the primitive recursion rule, which depends on knowing that \$h\$ is primitive recursive.Note well that what the primitive recursion rule demands as a premise is that \$h\$ is primitive recursive as a two-argument function. That is the function that describes how to combine \$n\$ and \$f(n)\$ in order to find the number you want to be \$f(n1)\$. In principle this \$h\$ needs to be applicable to every pair of numbers, not just ones where the second element happens to be \$f\$ applied to the first one. If you can not give such a general rule for \$h\$, the primitive recursion construction does not -- by definition -- necessarily produce a primitive recursive \$f\$.In response to the added material headed "edit": The construction you are quoting seems to arguing for the theorem that every constant function is primitive recursive. This conclusion is certainly true, but the argument you are quoting does not use the primitive recursion rule at all. It works by induction at the metalevel, but does not use recursion as a building block for functions.In fact the same argument would work to prove this:You should recognize rules 1-4 as exactly the same as the corresponding parts of the definition of primitive recursive functions. The primitive recursion rule is missing, but it should be clear that every "supersimple" function is necessarily also "primitive recursive".And the reason why the constant functions are supersimple is exactly the same as the reason why they are primitive recursive

3. Are all bijective functions continuous?

To the question in your title and last sentence: it is not true that all bijective functions are continuous.Consider the function from \$mathbbR\$ to \$mathbbR\$ (with the usual topology) given by \$\$f(x) = begincases x & text if x

ot in mathbbZ x 1 & text if x in mathbbZ endcases\$\$ Then this is a bijective function, sending integers to integers (and shifting them up by \$1\$) and sending all other real numbers to themselves. But it is not continuous.Generally, there is no reason to suspect a strong relationship between continuity and bijectivity.

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