Croom 5.4
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(a, b] and [c, d) are both homeomorphic to (0,1] which does not have the fixedpoint property: consider f (x) = x2 .
[a, ∞) is homeomorphic to (−∞, b]; both are homeomorphic to [0, ∞), which
trivially does not have the fixed point property: use f (x) = x + 1.
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b.No: f (x, y) = (x + 1, y + 1).
c. No: f (x, 0) = (0, x + 1) and f (0, y) = (0, y + 1) is a continuous function (even
at the origin!) that clearly has no fixed point.
d. Yes: homeomorphic to a closed interval in R.
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a. Trivial: increment each coordinate by 1.
b. Homeomorphic to Rn via the well-known “product tangent” mapping.
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Mendelson 4.4
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Pn
First we show that f (x) > 0 somewhere and f (x) < 0 somewhere. Let f (x) = i=0 ai xi
Pn−1
with n odd. Then f (x) = an xn (1 + i=0 aani xi−n ). Define bi = aani and notice
P
Pn−1
Pn−1
n−1
that | i=0 bi xi−n | ≤ i=0 |bi ||xi−n |. We wish to find when | i=0 bi xi−n | < 1,
so we bound each term on the RHS by n1 .
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If each |bi ||xi−n | <
1
n,
1
1
|x| > (n|bi |) n−i , so for |x| > (nmax(|bi |)) n−i , |
Pn−1
i=0
bi xi−n | < 1.
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Now, for x > (nmax(|bi |)) n−i , clearly f has the same sign as an .
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For x < −(nmax(|bi |)) n−i , f has the opposite sign of an since the xn term is
now negative.
Thus f has both signs and by IVT has a root (with absolute value less than or
equal to the bound discovered).
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Rb
f (t)dt
Note that by the IVT, f ([u, v]) ⊂ [U, V ]. a b−a is merely the average value
of
f on [a, b], which must of course lie between any lower and upper bounds, so
R
b
a
f (t)dt
b−a
∈ [U, V ] and such a w must exist in [u, v].
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Rt
We wish for Kf (t) = a F (x, f (x))dx = f (t). If f 0 (x) = F (x, f (x)) and
Rt
f (a) = 0 then clearly Kf (t) = a f 0 (x)dx = f (t) − f (a) = f (t).
Rt
Suppose Kf (t) = a F (x, f (x))dx = f (t). Then, taking the derivative w.r.t t
of both sides gives f 0 (t) = F (t, f (t)) . Plugging in gives f (t) − f (a) = f (t), so
f (a) = 0 as well.
Clearly S does not have the fixed point property (for any K with a “nice”
enough associated F ).
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