Vtyjkyuk

Let $a<b<c<d$ be real values, and let $f in C^infty([a,b])$ and $g in C^infty([c,d])$. Is there a way to "connect" these functions in a smooth way? That is, is there a function $h in C^infty([a,d])$ such that $h=f$ on $[a,b]$ and $h=g$ on $[c,d]$?

If this is true, how is the proven? Is the proof non-constructive, or is there an explicit way to do it?

There is a well known result of Borel that will be useful here.

Borel's theorem on power series: Let $x_0in mathbb R.$ Given $a_0,a_1, dots in mathbb R,$ there exists $fin C^infty(mathbb R)$ such that

$$tag 1 D^nf(x_0) = a_n, n=0,1,dots.$$

This is stated here, with a sketch of the proof, at the very end: http://math.caltech.edu/~nets/lecture12.pdf. The statement there is not exactly $(1),$ but the two are clearly equivalent. (I'd refer you to the wikipedia page on this, but the author there gets all caught up in stating a very general version, which we don't need.)

Corollary: If $fin C^infty[a,b],$ then there exists $Fin C^infty(mathbb R)$ such that $F= f$ on $[a,b].$

Proof: By Borel, there exists $f_1 in C^infty(mathbb R)$ such that

$$D^nf_1(a) = D^nf(a), n=0,1,dots.$$

There also exists $f_2 in C^infty(mathbb R)$ such that

$$D^nf_2(b) = D^nf(b), n=0,1,dots.$$

Now define

$$F(x) = begincases f_1(x) ,qquad x le a \ f(x) , qquad xin [a,b] \ f_2(x) ,qquad x ge b \
endcases$$

This $F$ does the job: It equals $f$ on $[a,b],$ it's clearly in $C^infty(mathbb Rsetminus a,b),$ and Borel gives us exactly the match-up at $a,b$ we need to see $Fin C^infty(mathbb R).$

So now to your set up with $fin C^infty([a,b]),gin C^infty([c,d]): $ We choose $F$ relative to $f,$ and $G$ relative to $g,$ as in the corollary. We also choose "blending" functions $alpha, beta in C^infty(mathbb R)$ that do the following:

$$alpha (x) = begincases 1, qquad xin (-infty,b] \ 0, qquad xin [c,infty)\
endcases$$

$$beta (x) = begincases 1, qquad xin [c,infty) \ 0, qquad xin (-infty,b]\
endcases$$

Then $h=alpha F + beta G$ solves your problem with room to spare: It belongs to $C^infty(mathbb R),$ it equals $f$ on $[a,b],$ and equals $g$ on $[c,d].$

If you extent $f(x)$ $forall epsilon>0$ from $[a,b]$ to $[a,b+epsilon]$ (which I guess can be done in the sense of the series-expression as all derivatives are defined), so that you are free to do whatever you want on $(b,b+epsilon]$ and likewise $g(x)$ from $[c,d]$ to $[c-epsilon,d]$, then you can blend it by
$$
h(x) = leftThetaleft[ a leq x leq b right] + varphi_epsilonleft(x-bright)right f(x) + leftThetaleft[ c leq x leq d right] + varphi_epsilonleft(c-xright)right g(x)
$$
where
$$
Thetaleft[ a leq x leq b right] = begincases 1 qquad a leq x leq b \ 0 qquad rm else endcases
$$
and
$$
varphi_epsilonleft(xright) = begincases 0 qquad x leq 0 \ tanh^2left(frac1xcoshleft(frac1x-epsilonright)right) qquad 0 < x < epsilon \ 0 qquad epsilon leq xendcases , .
$$