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Prove that if x <= c <= y, then c = ax+by, for some a,b>=0,
a+b=1.

a = (y-c)/(y-x)

b = (c-x)/(y-x)

b = (c-x)/(y-x)

Show a function which is defined on the real line and
continuous at only one point.

0 for rationals,

x for irrationals

continuous only at 0

x for irrationals

continuous only at 0

Show a function which is defined on the real line and
differentiable only at one point.

0 for rationals,

x*x for irrationals

differentiable only at 0, because continuous only at 0

x*x for irrationals

differentiable only at 0, because continuous only at 0

Sketch the proof of Rolle's theorem.

First prove that the function is either constant or has a local
extremum.

Later use the fact that if a differentiable function has a local extremum at a point,

then its derivative is equal to zero at that point.

page 154 in Golden Gate

Later use the fact that if a differentiable function has a local extremum at a point,

then its derivative is equal to zero at that point.

page 154 in Golden Gate

State and prove Cauchy's Mean-value Theorem.

You can refer to Rolle's theorem.

You can refer to Rolle's theorem.

page 155 in Golden Gate

Consider a real-valued differentiable function defined on a
bounded closed interval. Does it have to be a Lipschitz function?

NO. g:[0,1]->|R, g(0) = 0, g(x) = x*x*sin(exp(1/x))

This function is differentiable, g'(0) = 0, but g' is not bounded. Hence it is not Lipschitz.

Put x[n]=1/log(n*2*pi). Show that lim(n->oo) g'(x[n]) = -oo.

Thus the derivative is not bounded.

This function is differentiable, g'(0) = 0, but g' is not bounded. Hence it is not Lipschitz.

Put x[n]=1/log(n*2*pi). Show that lim(n->oo) g'(x[n]) = -oo.

Thus the derivative is not bounded.

Let f be a real-valued function defined on an interval on the
real line. Suppose that it is monotonic and that its image is
an interval. Does the function have to be continuous?

YES.

Use the fact that a bounded monotonic function has a limit.

Use the fact that a bounded monotonic function has a limit.

Let f be a function from a metric space, say X, to a complete
metric space. Let A be a nonclosed subset of X. Let f be a
uniformly continuous function on A. Can f be extended to the
closure of A so that the extended function is continuous?

YES. And what's more, the extension is uniformly continuous.

Completeness is essential. In the proof, we use the Cauchy criterion for the existence of the limit of a function at a point.

Completeness is essential. In the proof, we use the Cauchy criterion for the existence of the limit of a function at a point.

Consider a function from one metric space into another. Suppose
it is continuous and uniformly continuous on a dense set. Does
this function have to be uniformly continuous?

YES. The proof can be performed routinely.

(1) f : R -> R

(2) f has n derivatives

(3) the n-th derivative is constantly zero

What can we infer about this function?

(2) f has n derivatives

(3) the n-th derivative is constantly zero

What can we infer about this function?

It's a polynomial of degree less than n.

We can use Taylor's formula for an easy proof, or we can prove it by induction.

We can use Taylor's formula for an easy proof, or we can prove it by induction.