The article presents a new proof of the beautiful and difficult theorem of mathematical analysis, presented in such a way that it is accessible to students of the senior classes of specialized mathematical schools.

Let be $$$f (x)$- infinitely many times differentiable real function, and for each point $$$x \ in R$there is a natural $$$n$such that $$$f ^ {(n)} (x) = 0$. Then $$$f (x)$polynomial.

Evidence

We need the Baire theorem on a system of closed sets:

1. Let $$$H$and $$$F_ {1}, F_ {2}, ..., F_ {n}, ...$closed subsets of a line, and $$$H \ neq \ varnothing$and $$$H \ subset \ bigcup \ limits_ {n} F_ {n}$. Then in $$$H$there is a point which is contained in one of $$$F_ {n}$along with your neighborhood. More precisely, there is a point $$$x \ in H$, natural $$$n$and $$$\ varepsilon> 0$such that $$$(x- \ varepsilon; x + \ varepsilon) \ cap H \ subset F_ {n}$.

Indeed (by contradiction), choose a point $$$x_ {1} \ in H$and surround it with a neighborhood $$$\ Delta_ {1} = (x- \ varepsilon_ {1}; x + \ varepsilon_ {1})$where $$$\ varepsilon_ {1} <1$. We assumed that the statement of the Baer theorem is not true. Means $$$\ Delta_ {1} \ cap H \ not \ subset F_ {1}$. Pick in $$$\ Delta_ {1} \ cap H$the point $$$x_ {2} \ notin F_ {1}$. Surround $$$x_ {2}$interval $$$\ Delta_ {2} = (x_ {2} - \ varepsilon_ {2}; x_ {2} + \ varepsilon_ {2})$such that the ends of this interval are points $$$x_ {2} - \ varepsilon_ {2}$and $$$x_ {2} + \ varepsilon_ {2}$lie in $$$\ Delta_ {1}$, but $$$\ varepsilon_ {2} <\ frac {1} {2}$. By assumption $$$\ Delta_ {2} \ cap H \ notin F_ {2}$. This allows you to choose in $$$\ Delta_ {2} \ cap H$some point $$$x_ {3} \ notin F_ {2}, ...$Continuing the process, we will build a nested tightening sequence of intervals. $$$\ Delta_ {1} \ supset \ Delta_ {2} \ supset ...$It's clear that
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$$$x_ {1} - \ varepsilon_ {1} <x_ {2} - \ varepsilon_ {2} <... <x_ {n} - \ varepsilon_ {n} ...$, (one)
$$$x_ {1} + \ varepsilon_ {1}> x_ {2} + \ varepsilon_ {2}> ...> x_ {n} + \ varepsilon_ {n} ...$(2)

Since every gap $$$\ Delta_ {i} \ cap H \ neq \ varnothing$then $$$\ lim _ {i \ to \ infty} (x_ {i} - \ varepsilon_ {i}) = \ lim_ {i \ to \ infty} (x_ {i} + \ varepsilon_ {i}) = y, y \ in H$, and from (1) and (2) it follows that $$$y \ in \ Delta_ {i}$for each $$$i$. So we found the point $$$y \ in H$but not in any of the sets
$$$F_ {i} \ phantom {1} (i = 1,2, ...)$.

We say that a point on a real line is correct if, in a certain neighborhood of this point, the function $$$f (x)$- polynomial. The set of all regular points is denoted by $$$E$. Lots of $$$E '$additional to $$$E$denote by $$$F$and call the set of irregular points. (Let's say that if $$$x \ in F$then $$$x$- wrong point).

2. If each point of the segment $$$[a; b]$correct then narrowing $$$f (x)$on $$$[a; b]$- polynomial.

Indeed, for each point $$$t \ in [a; b]$there is an interval such that the narrowing $$$f (x)$this interval is a polynomial. Those. for each point there is an interval and some kind of natural $$$n$, what $$$f ^ {(n)} (x)$equals zero on this interval.

From the compactness of the segment $$$[a; b]$it follows that there is such a natural $$$m$, what $$$f ^ {(m)} (x) = 0$everywhere on $$$[a; b]$, Consequently $$$f (x)$- polynomial.

3. If each point of the interval $$$[a; b)$correct then
constriction $$$f (x)$on $$$[a; b)$- polynomial.

Evidence. Consider the increasing sequence $$$a = a_ {1} <a_ {2} <a_ {3} <... <a_ {n} <...$such that $$$a_ {n}$converges to $$$b$. As proved in the previous paragraph on each of the segments $$$[a_ {1}; a_ {2}], ..., [a_ {1}; a_ {n}], ...$constriction $$$f (x)$- polynomial. Let be $$$P_ {k} (x)$- polynomial coinciding with $$$f (x)$on the segment $$$[a_ {1}; a_ {k + 1}]$. It's clear that $$$P_ {k} (x) = P_ {1} (x)$for all $$$k = 2,3, ...$therefore $$$P_ {1} (x)$matches with $$$f (x)$on $$$[a; b)$means at the point $$$b$. (Recall that $$$P_ {1} (x)$and $$$f (x)$continuous everywhere on $$$R$).

Similar to the previous one, it is easy to prove that:

4. If each point of the interval $$$(a; b]$or interval $$$(a; b)$- correct, then $$$f (x)$- polynomial on $$$[a; b]$.

We proceed to the study of irregular points, i.e. set points $$$F$.

5. The set $$$F$does not contain isolated points.

Really. Let be $$$a \ in F$- isolated point. Then for some $$$\ varepsilon> 0 \ phantom {1} [a- \ varepsilon, a)$and $$$(a, a + \ varepsilon]$consist of correct points. Mean narrowing $$$f (x)$on $$$[a- \ varepsilon; a]$and on $$$[a; a + \ varepsilon]$polynomials. It is clear that with a sufficiently large $$$n$( $$$n$there must be more degrees of each of these polynomials) $$$f ^ {n} (x)$will be zero everywhere on $$$[a- \ varepsilon; a + \ varepsilon]$. Those. $$$a$is the right point.

6. Let the set $$$F$wrong points is not empty. Set $$$E_ {n} = \ {x: f ^ {(n)} (x) = 0 \}$ . It's clear that $$$F \ subset \ bigcup \ limits_ {n} E_ {n}$and each $$$E_ {n}$is closed. From the Baire theorem (see 1.) it follows that there exists an interval $$$(a; b)$such that $$$(a; b) \ cap F \ neq 0$and $$$(a; b) \ cap F$lies in one of $$$E_ {n}$.

Consider the function $$$f ^ {(n)} (x)$. This function is zero at every point. $$$x \ in F \ cap (a; b)$. Since every irregular point is the limit for the set $$$F$then $$$f ^ {(n + k)} (x) = 0$for all wholes $$$k \ geq 0$and all $$$x \ in (a; b) \ cap F$.

Prove that $$$f ^ {n} (x)$equals $$$0$everywhere on $$$(a; b)$. Let not so. Then there is $$$c \ in (a; b)$such that $$$f ^ {(n)} (c) \ neq 0$. Since many $$$F$not empty and closed, then we find a point in it $$$d$closest to $$$c$. For definiteness, we set $$$d <c$. Function $$$g (x) = f ^ {(n)} (x)$infinitely many times differentiable on $$$[d; c], \ phantom {1} g (d) = 0$and all derivatives $$$g ^ {n} (d) = 0$. Because $$$g (c) \ neq 0$then by the theorem on finite Lagrange increments $$$g ^ {(n)} (x)$can't be zero everywhere on $$$(d; c)$not for one natural $$$n$.

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Slobodnik Semen Grigorievich ,
content developer for the application "Tutor: Mathematics" (see the article on Habré ), Ph.D. in Physics and Mathematics, teacher of school mathematics 179 Moscow