Theorems on the continuous functions
1) If a function f is continuous in x0 ? X and f(x0) > 0 exists around a 0 U ofx such that f(x) > 0 for every x ? U
2) It is x0 ? To and f(x0) ? B and if f it is continuous in x0 and g is continuous in f(x0) then the composed function g°f is continuous in x0
3) If a function is monotonous on an interval then f it can have to infinite more a countable one than points of discontinuity, and are only of 1ª specie(o dismissable if they are found to the ends of the interval).
For a function monotonous the limit exists always and
coincides with the sup or the inf to second of the cases therefore if
we have one point x0 they will
have to exist ended both the 
limits and 
, if they are equal the function are continuous while if is
various the function introduces one jump discontinuity.
It remains to demonstrate that it knows to you possible are countable infinity that is that they represent an entirety that can be placed in correspondence biunivoca with the entirety ?, to such need it is observed that if we consider it knows to you greater of 1/n with n ? ? they are countable infinity for every value that we give to n and siccome ? same he is countable achieves some that totally we will have one countable infinity of discontinuity of 1ª species or jump.
4) Theorem of the zeroes :
Description: If a defined continuous function on an interval [ a,b ] assumes to its extreme not null values and of opposite sign, then f admits at least one zero y ? (a,b).
Supposing that f(a) < 0 while f(b) > 0 the demonstration of the existence at least a zero happens with a dicotomico procedure characterized from the following steps:
a) individual medium point c of the interval puts into effect them
b) based on the value assumed from the function in c I execute one of the following steps:
* f(c) = 0 I have found the zero. Algorithm completed with happening.
* f(c) < 0 the medium point puts into effect them becomes the new left edge of the interval to consider.
* f(c) > 0 the medium point puts into effect them becomes the new skillful edge of the interval to consider.
c) is returned to the point to).
In such a way 2 successions have been created, a {ton} of the left edges and {bn}of the edges rights, it must be
demonstrated that both converge to the same point y and that in this
point the function is worth 0, { nin particular} being a succession monotonous and limited converges to a
point y while {bn} converges to the
same point y in how much the distance between the terms n of the
succession is equal 
a that stretches to 0 for n ® ¥. Moreover the procedure of
construction of the successions was such that f{ton} £ 0 while f{bn} ³ 0 for n ® ¥. It is only the continuity of
the function to assure that the two limits for n ® ¥must be equal and therefore f(y) = 0 to us.
5) If 2 continuous functions f and g defined on the same interval [ a,b ] are such for which to the skillful end f he is greater of b while to the left end g he is greater of f exists at least one solution y ? (a,b) of the equation f(x) = g(x).
The theorem of the zeroes is applied to the continuous function f(x) - g(x).
6) a continuous function on an interval, not necessarily limited, assumes all the values comprised between sup and inf. Analogous it can be said that the image of a defined continuous function on an interval, is same it an interval.
For every y comprised between sup and the inf I apply the theorem of the zeroes to the function f(x) - y on a chosen interval [ a,b ] so that the sign of the function in to is various from the sign of the function in b.
7) If a function f is continuous and invertible on an interval f it is closely monotonous .
It is demonstrated for absurdity supposing that the function is not closely monotonous and taking advantage of the continuity is succeeded in to demonstrate that the function is not iniettiva and therefore is not invertible, remembering that tightened monotonia implies that if in the dominion x < is had then y < z in the image it has f(x) < f(y) < f(z) while for iniettività agrees that however presi x, y, z has f(x) ¹ f(y) ¹ f(z). If we suppose for absurdity that f(y) he is greater of the others 2 values, it is observed that in the interval between y and the point correspondent to the minor between f(x) and f(z) as an example z (interval [ y,z ]) the function assumes all the values comprised between inf and sup between the which the therefore also f(x) therefore function it is not inettiva, that confirmation the validity of the theorem.
8) If f it is a continuous and invertible function then also the inverse one is continuous in its dominion if:
a) the dominion of f is an interval
b) the dominion of f is a closed and limited entirety (compact).
Considering the case of an interval:
If a function is continuous and invertible on an interval
then for 51) the such function it is closely monotonous therefore also the inverse function must closely monotonous nonchè be defined on an interval in how much image
of a continuous function on an interval. From the property of
the functions monotone the function is deduced that choseny 0 must exist 
ended 
e, affinchè turns out
continuous these 2 limits must be equal otherwise would be a jump, but
if the image of f -1 were a jump then that is the dominion of f would not be an
interval against the hypothesis of the theorem.
9) If I then have a continuous function on an interval [ a,b ]:
a) f is limited in [ a,b ]
b) exists the Max and the Min on [ a,b ]
c) the image of f is comprised between the Max and the Min.
The theorem is based on 3 hypotheses that cannot be weakened that is f must be a function continuous, limited (that is racchiudibile in around of the origin) and defined on a closed interval.
10) If I have a defined continuous function on a compact K
a) the image f(K) is also it a compact one.
b) if K " then exists the Min and the Max of f in f(K).
to) the compactness of f(K) it implies that, taken a
succession yK in f(K) a
subsuccession y LK must exist that stretches to y0 ? f(K)
for k® ¥ , to such aim it can be observed that
such succession yK is image of
a succession xK on K for which, being K a compact
one, exists one subsuccession xLK that stretches to x0 ? K for k® ¥.
Is la continuity of f in x0 assuring to us that 
therefore its image, subsuccession yLK ,
stretches to f(x0) and f(K) it
is therefore a compact one.
b) as soon as we have demonstrated that f(K) it is compact that is a closed and limited entirety. From the limitatezza it is deduced for Bolzano-Weierstrass that must exist the inf and the sup while remembering that a closed entirety contains also its frontier of it it derives that the Sup is equivalent to the Max and the Inf is equivalent to the Min.
11) If I have a continuous and invertible function on compact then an inverse one is continuous in its dominion f(K).
It is necessary to demonstrate that a succession yK to values in f(K) that it stretches to y?f(K) (y it must be stung not isolated in how much if isolated the inverse one is continuous in y for same definition of continuity in a point) has as image one succession xK which converges to x = f -1(y). One supposes for absurdity that xK does not stretch to f -1(y) that is that its subsuccession x LK existsone that does not converge to f -1(y) (is remembered that a succession has every limit l > its subsuccession has limit l), from this per² finding to us on a compact one can be extracted one subsuccession xJK which converges to x ? K. The continuity of f in x assures to us that the image of succession xJK stretches to f(x) and siccome the limit is only then x = f -1(y). To this point it is had that the subsuccession converges to f -1(y) that it is impossible therefore leaving from the absurd hypothesis that xK does not stretch to f -1(y) is reached the absurdity that one its subsuccession instead stretches to f -1(y) that it is impossible inasmuch as the values of the subsuccession are extracted from the main succession.
12) Theorem of Heine - Cantor
Description: a continuous function on a compact one is also uniform continuous.
It is proceeded for absurdity denying that the function is uniform continuous that is exists and at least0 > such 0 such that for every d > 0 are 2 pointsx d andy d that they to be distant between of they less than d but their images to be distant between of they more than and0. Giving to d of values 1, 1/2..., 1/k 2 successions x Kand y Kare obtained constituted from points that deny the uniform continuity of the function. Finding to us but on a compact one pu² to extract from xK one subsuccession xLK which converges to an element x. Of rest but also yLK it converges to x for KL ® ¥ in how much elements KL of the 2 sottosuccessioni to be distant between of they less than 1/KL Dato that both the sottosuccessioni stretch to x, and for the continuity of f, follows that the distance between the 2 sottosuccessioni stretches to zero to the contrary of how much before asserted, leaving therefore gives task that the f is not uniform continuous reaches the absurdity that the subsuccession respects the uniform continuity while the succession from which it is extracted not respects it.
13-a) If f and g they are 2 uniform continuous functions then also f ± g and to* f, they are uniform continuous functions.
For the product instead it is not true.
13-b) If g it is a uniform continuous function in X and f it is one uniform continuous function in g(X) then also f°g is a uniform continuous function in X.
The definition of uniform continuity of g°f is based on the fact that the distance between the images of 2 points x and y pertaining to X is smaller of and and the distance between g(x) and g(y) is smaller of dthe 1 function of and that it respects the definition of uniform continuity of f, and moreover the distance between x and y is smaller of dthe 2 which it is function of dthe 1 that I repeat is function of and.
14) If f and a function uniform continues on X f is estendibile with continuity
R-alla.chiusura of X, one function exists 
that
is that is the extension of f R-alla.chiusura of X that is to 
.
The definition of uniform continuity implies the existence
of an ended limit and therefore does not have to
make other that to replace to end x0 in the f which is not defined the limit of the f for x ® x0 .
15) If a function f is uniform continuous in X it it is also in every sottoset of X
16) If X are one with limited then
the
function f is uniform continuous > f is estendibile with continuity to .
already is demonstrated to the point 58)
?
It is uniform continuous in how much
continuous one on the compact that for 59) it
implies that is uniform continues
also on X and inasmuch as on X coincides with f,
follows that f it is uniform continuous.
17) If f it is uniform continuous in X then f it is limited in every sottoset limited of X.
The function f is uniform continuous also in every
sottoset To of X, and therefore R-alla.chiusura of To can be
extended with continuity that is 
on which but is limited in how much for
a continuous function on a compact one exists Max and Min, and therefore since on To coincides
with f of it achieves that f it is limited on To.
18) If f it is uniform continuous in X and the interval [ b, ¥) X then exists To, such B ³ 0 that |f(x)| £ Ax B for every x ³ b.
For symmetry he is sufficient to demonstrate that such A,B ³exist 0 that f(x) £ every Ax B for x ³ b.
Being the continuous function uniform, placing and = 1, it is had that for every brace x,y following to a sure point b the inequality is verified |f(x)-f(y)| < 1.
Replacing in this in place of y the point b and disuguagliando through Cauchy-Schwarz it is had that to the end of the interval, that is in the point b d, it is f(b d) £ 1 |f(b)|.
If instead it is replaced to y the point b d and si disuguaglia through Cauchy-Schwarz it has that to the end of 2° the interval, that is in the point b 2d, it is f(b 2d) £ 2 |f(b)| .
Continuing and joining together the values of the function to the ends of these intervals a straight one is obtained that from b in then all is found over the function.
19) In the case that f is a continuous function on with limitless then affinchè it is also uniform continuous must uniform continuous in every sottoset be limited of X and must be verified one of the 3 following conditions.
a) X is limited inferiorly (advancedly) and f it has a horizontal or oblique asymptote for x® ¥.
b) X is limited inferiorly (advancedly) and exists a such R that in (R, ¥) f is lipschitziana.
c) f is periodic in X.
where for limited inferiorly one agrees that with of definition of the function it does not comprise -¥.