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Proofs of big theorems of calculus

  1. Mar 4, 2007 #1

    mathwonk

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    The following theorems are usually left unproved in calculus, for no good reason. See what you think.

    2250: Elementary proofs of big theorems
    The first theoretical result is the Intermediate Value Theorem (IVT) for continuous functions on an interval.
    Theorem: If f is continuous on then interval I, then the set of values f assumes on I is also an interval. I.e. if a,b,are points in I, then any number between f(a) and f(b) is also a value of f, taken at some point between a and b.
    proof: We assume f(a) < 0 and f(b) >0, and try to find c with f(c) = 0. We assume that every infinite decimal represents a real number.

    Lemma: If f is continuous at c and f changes sign on every interval containing c, then f(c) = 0.
    proof: This is the contrapositive of the fact that if f is positive (or negative) at c and continuous at c, then f is positive (or negative) on some interval containing c. ("Sign preserving property", p. 114, #60.) QED.

    Thus it suffices to find a real number c such that f changes sign on every interval containing c. Assume [a,b] = [0,1].
    Since f(0) < 0 and f(1) >0, then f changes sign on some interval of form [r/10, (r+1)/10]. let c start out as the decimal .r.
    Then since f(r/10) < 0 and f((r+1)/10) > 0, f changes sign on some interval of form [(10r+s)/100, (10r+s+1)/100]. Then c continues as the decimal .rs.
    Continuing in this way forever, we obtain an infinite decimal, i.e. a real number c = .rs....., in the intyerval [0,1], such that f changes sign on every interval containing c. Hence f(c) = 0. QED.

    We know the continuous image of an open bounded interval may be neither open nor bounded. But the next big theorem says that the continuous image of a closed bounded interval is also closed and bounded. We do it in two steps:
    Theorem: If f is a function which is continuous everywhere on a closed bounded interval [a,b], then f is bounded there.
    proof: We prove it by contradiction, i.e. assuming f is unbounded leads to finding a point whewre f is not continuous.

    Lemma: A function which is continuous at c, is also bounded on some interval containing c.
    proof: This is immediate from the definition of continuity. E.g. if e = 1, by continuity of f at c, there is an interval I containing c where the values of f lie between f(c)-1 and f(c)+1. Thus f is bounded on I. QED.

    Hence it suffices to show that if f is unbounded on [a,b], then there is a point c of [a,b] such that f is unbounded on every interval containing c.
    Assume [a,b] = [0,1], and that f is unbounded on [0,1]. Then there is some interval of form [r/10, (r+1)/10] where f is unbounded. Start out the decimal c as .r.
    Then there is some interval of form [(10r+s)/100, (10r+s+1)/100] where f is unbounded. Continue the expansion of c as the decimal .rs.
    Continuing forever, we construct an infinite decimal c = .rs......, in the interval [0,1], such that f is unbounded on every interval containing c. Thus f is not continuous at c. QED.

    Theorem: If f is continuous on the closed bounded interval [a,b], then f assumes a maximum value there.
    proof:
    We know f has an upper bound on [a,b]. We want a smallest one.
    Lemma: A non empty bounded set has a smallest upper bound c.
    proof: If S is a non empty bounded subset say of [0,1], in particular 1 is an upper bound for S, so there is some number of form r/10 which is not an upper bound of S, but such that (r+1)/10 is an upper bound. Let the decimal c start out as .r.
    Then there is some number of form 10r+s which is not an upper bound but such that (10r+s+1)/100 is an upper bound. Continue c as .rs.
    Continuing forever we construct a real number c = .rs...., such that no smaller number is an upper bound, but every larger number is. Hence c is the smallest upper bound for S. QED.

    Now if K is the smallest upper bound of all the values of f on [a,b], it will suffice to show f assumes the value K on [a,b]. But if not, then g(x) = 1/(K-f(x)) would be both continuous and unbounded on [a,b], which contradicts the previous result. QED.

    A similar criterion for open intervals is this:
    Cor: If f is continuous on (a,b), and f(x) approaches plus infinity as x approaches a, and also as x approaches b, then f has a global minimum on (a,b). (Apply the previous result to a suitable closed interval of form [a-e, b-e].)
    For more general max min problems on open intervals it, helps to have a little more understanding of how the derivative of a function affects the behavior of the graph . For some reason the following simple principle seems not to be stated in standard books.
    Theorem: A continuous function on an interval cannot change direction except at a critical point. I.e. f is strictly monotone on any interval not containing a critical point.
    proof: If has no critical points on an interval, then f is not constant. If f is not monotone then the IVT implies that f takes the same value twice on that interval, say f(a) = f(b) for some points a < b in the interval. Then f has both a maximum and a minimum on the interval [a,b] and since f is not constant, one of these extrema occurs at some c with a < c < b. Then c must be a critical point, as we know. Since there are no critical points, in fact f is monotone. QED.

    This gives us many easily verifiable criteria for finding maxima and minima on open intervals. Here is a typical one.
    Cor: If f is differentiable on (a,b) and has only one critical point c in that interval, then f(c) is a global minimum for f on (a,b) provided there are points u, v, with a< u< c < v < b, i.e. u and v are on either side of the critical point c, and f(c) < f(u) and f(c) < f(v).
    proof: f is monotone on each side of c, and greater somewhere hence greater everywhere. QED.

    More generally if f goes up somewhere on the left of the first critical point, and also goes up somewhere on the right of the last one, then f has a global minimum at some critical point.

    Finally we have:
    proof: By the argument in the proof of the previous theorem, if a differentiable function h takes the same value at two different points, then h’ = 0 somewhere in between. Applying this to a difference h = f-g, shows two functions taking the same value at two points must have the same derivative somewhere in between. Since for any a,b in I, f agrees at a and b with the linear function passing through (a,f(a)), (b,f(b)), whose slope is everywhere equal to ([f(b)-f(a)]/[b-a]), the graph of f must have that same slope at some c between a and b. In particular if f(a) differs from f(b), there is a c with a < c < b and f’(c) = ([f(b)-f(a)]/[b-a]), which is not zero. QED.

    Cor: If f, g have the same derivative everywhere on an interval, then f differs from g by a constant on that interval.
    proof: Since (f-g)' is zero everywhere, (f-g) must be constant. QED.

    Cor: If f has no critical points on an interval (a,b) and if f'(c) > 0 at some point c of (a,b), then f is strictly increasing on (a,b).
    proof: By definition of the derivative, f is increasing at c, and f is monotone on (a,b), so f is increasing everywhere on (a,b). QED.
     
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  3. Mar 5, 2007 #2

    Gib Z

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    Yea I just noticed that as well, its always stated in textbooks but never proved...I don't see why they should be left unproved, MVT is quite elementary, and best for newbies to calc, intuitive.
     
  4. Mar 5, 2007 #3
    I guess it is the intuitive content of the theorem that leads to it being taught without proof. IMO the proof of MVT is not interesting, but the uses of MVT are.
     
  5. Mar 5, 2007 #4

    JasonRox

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    Very true.

    But leaving proofs out gets students all frustrated when it comes to proving things that are even more intuitive than the MVT. I think the MVT is a great place to start for constructing proofs.
     
  6. Mar 5, 2007 #5
    Mathwonk, I'm not really sure what the students this is adressing really already know. So I'll simply add one remark: At the time I found one particular proof of the IVT quite enlightening. For several reasons: It's easy to remember, it shows what you really need for the IVT (i.e. the real numbers and some ideas on continuity), which I think is immensely helpful.

    I think the proof is well known, so I'll keep it short: You take the interval [itex]I:=[a,b][/itex], assume [itex]f(a)>0[/itex] and [itex]f(b)<0[/itex]. Then you look at [itex]f((a+b)/2)[/itex], the value of f at the midpoint. If that's =0, you're done, if it's >0, you set [itex]I_1:=[(a+b)/2,b][/itex] and repeat, if it's <0 you set [itex]I_1:=[a, (a+b)/2][/itex] and repeat the procedure. The length of the intervals [itex]I_k[/itex] obtained in that manner will decrease as [itex]2^{-k}[/itex] and you get, thanks to the properties of the reals, one point as limit. This easily generalizes to any value of the function in between [itex]f(a)[/itex] and [itex]f(b)[/itex].
    So: what do you think, mathwonk?
     
  7. Mar 6, 2007 #6

    mathwonk

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    i like it. it is the same as mine above, except that instead of my taking as an axiom that there is a point in all those intervals, i actually construct the point as a decimal. i.e. students already believe that an infinited ecimal is a real number, whereas in the simpelr construction above they would need to believe an infinite binary expansion is a real number.

    also note that in my exposition i never mention the MVT anywhere. I agree with you guys that the uses are morer interesting thena the statement so i never give the statement, instead my theorems are stated in etrms of the interesting consequences instead of the uninteresting usual MVT statement.

    it is silly to ask students tor ecall the statement of theMVT if you really want them to recall the corollaries. so i made the corollaries the statement.

    note alkso that my version is stronger than the usual MVT since I get that a function is monotone on any interval not containing a critical point. in the usual MVT you have to check that the sign of the derivative does not change on the interva. in my version that is a consequence of the more pr4ecise form of the argument.

    (And i do not assume continuity of the derivative. i.e. i am proving ALL derivatives have the intermediate valoue proeprty whetehr they are continuious or not. and note a derivative of a differebntiable function does not need to be continuous even though the fucntion is. e.g.x^2 sin(1/x) is differentiable also at 0, if you set the value to zero there, but its derivative is 2xsin(1/x)- cos(1/x) which is not even continuous at zero.

    still that derivative does have the intermediate value property, as you will see by graphing it, or trying hard enough.
     
  8. Mar 6, 2007 #7

    mathwonk

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    notice also that instead of assuming a non empty bunded set has a least upper bound, i prove it by again exhibiting the lub as a decimal.
     
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