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I am working on exercice 9.5.2 of Analysis by Zorich and I am stuck at the question b.

a) A set $Esubset X$ of a metric space $(X,d)$ is nowhere dense in X if it is not dense in any ball, that is, if for every ball $B(x,r)$ there is a second ball $B(x_1,r_1)subset B(x,r)$ containing no points of the set $E$.
A set E is of first category in X if it can be represented as a countable union of nowhere dense sets. A set that is not of first category is of second category in $X$. Show that a complete metric space is set of second category (in itself).

b) Show that if a function $fin C^{(infty)}[a,b]$ is such that $forall xin [a,b] ;exists nin mathbb{N} ;forall m>n ;(f^{(m)}(x)=0)$, then $f$ is a polynomial.

Here's my try: Define the sets $S_n:={xin[a,b]mid f^{(m)}(x)=0; forall m>n}$. Then $cup_{n=1}^{infty} S_n =[a,b]$. As $[a,b]$ is a complete metric space, $S_n$ cannot be all nowhere dense. Define $Y:={xin [a,b] mid text{ there exists a neighborhood of } x text{ and } n text{ such that } S_n text{ is dense in that neighborhood}$. I want to say that $Y=[a,b]$ so that I could conclude with the compactness of $[a,b]$. But all I can say is that the complement of $Y$ in $[a,b]$ contains no interval.

Can you help me? Thanks!

Here is a sketch of a possible approach, which may have some holes, but it is too long for a comment. Maybe someone can patch it up.Picking up on your idea, set

$T = {tin [a,b]: forall (c,d)ni t: frestriction_{(c,d)}$ is not a polynomial$}$

Then $T$ is non-empty and closed. By construction, it has no isolated points. Now we may apply the Baire theorem on ${Tcap S_n}$ to find an interval $(c,d)$ such that for some $nin mathbb N, (c,d)cap Tsubset S_n$.

Now, $f$ is a polynomial on $(c,d)setminus T,$ which is open and so contains an interval $(alpha,beta)$, which we may take to be maximal: indeed, $(c,d)setminus T$ is a countable disjoint union of intervals $bigcup_n(a_n,b_n)$ and so ${x:a',b'in (a,b)setminus T text{and} b'-a'ge b_n-a_n}$ is a maximal interval in $(a,b)setminus T$.

Now, either $alpha$ or $betain T$. Suppose $alphain T$. Then, on every interval $(c',c'')$ containing $alpha, f$ is $not$ a polynomial. Choose $c'''$ such that $c'<alpha<c'''<c''<d$. Then, $f$ is not a polynomial on $(c''',c'').$ But since $(c''',c'')subseteq (alpha,beta)$, so we have a contradiction.

So, either there is no such interval $(c,d)$ or $T$ is empty. In either case, the result follows.

You can ‘trace’ the $m$th derivative back to the original function recursively. If $f^{(m)}(x) = 0$ for all $x in [a,b]$, then $f^{(m-1)}(x) = int f^{(m)}(x) , mathrm{d}x = int 0,mathrm{d}x=c_0$. Subsequently, $f^{(m-2)}(x) = int c_0,mathrm{d}x=c_1x+c_0$. Applying this technique $m$ times, we see that $f(x) = c_mx^m + c_{m-1}x^{m-1}+dots + c_1x + c_0$, which is a polynomial by definition.

(Note that the indices $c_1, c_0$ etc. are not necessarily the same for each iteration, they are just arbitrary constants, where the index indicates the power of $x$ associated to it.)

b) Show that if a function $fin C^{(infty)}[a,b]$ is such that $forall xin [a,b] ;exists nin mathbb{N} ;forall m>n ;(f^{(m)}(x)=0)$, then $f$ is a polynomial.

It suffices to require that $f^n(x)=0$ and this is a well-known fact.