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Prove that for each non-closed subset of |R, there exists a
continuous function defined on this set, which is not uniformly
continuous.

If p is the point belonging to the closure of the set but not
to the set itself, put f(x) = log(|x-p|).

page 185 in GOLDEN GATE

page 185 in GOLDEN GATE

Prove that for each non-closed subset of |R, there exists a
bounded continuous function, which has no maximum.

If p is the point belonging to the closure of the set but not
to the set itself, put f(x)=1/(1+|x-p|).

(1) g : (0,oo) -> |R is diffable

(2) lim(x->oo) g(x)+g'(x) = A; A:-|R*

What can we conclude from this?

(2) lim(x->oo) g(x)+g'(x) = A; A:-|R*

What can we conclude from this?

(3) lim(x->oo) g(x) = A

Hint: use the exponential function and L'Hospital's Rule.

page 190 in golden gate

Hint: use the exponential function and L'Hospital's Rule.

page 190 in golden gate

Let f be a real-valued function defined on an interval. We
informally say that f is convex iff for every two points x,y
belonging to the domain of the function the segment joining
points (x,f(x)) and (y,(f(y)) lies above the graph of the
function.

Provide two formalizations of this idea.

Provide two formalizations of this idea.

(1) for every x,y, every a,b>=0 such that a+b=1

f(ax+by)<=af(x) + bf(y)

(2) for every distinct x,y, for every x<c<y,

f(c) <= ((f(y)-f(x))/(y-x)) * (c-x) + f(x)

f(ax+by)<=af(x) + bf(y)

(2) for every distinct x,y, for every x<c<y,

f(c) <= ((f(y)-f(x))/(y-x)) * (c-x) + f(x)

f : I -> R, I is an interval on the real line

(1) for every x,y, every a,b>=0 such that a+b=1

f(ax+by)<=af(x) + bf(y)

(2) for every distinct x,y, for every x<=c<=y or y<=c<=x

f(c) <= ((f(y)-f(x))/(y-x)) * (c-x) + f(x)

(1) for every x,y, every a,b>=0 such that a+b=1

f(ax+by)<=af(x) + bf(y)

(2) for every distinct x,y, for every x<=c<=y or y<=c<=x

f(c) <= ((f(y)-f(x))/(y-x)) * (c-x) + f(x)

Hint:

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

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

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

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

f:I->|R; I is an interval on the real line

(1) /\(x,y:-I) /\(a,b>=0, a+b=1) f(ax+by) <= af(x) + bf(y)

(2) /\(x<c<y :- I) f(c) <= ((f(y)-f(x))/(y-x))*(c-x) + f(x)

Prove that (2) => (1)

(1) /\(x,y:-I) /\(a,b>=0, a+b=1) f(ax+by) <= af(x) + bf(y)

(2) /\(x<c<y :- I) f(c) <= ((f(y)-f(x))/(y-x))*(c-x) + f(x)

Prove that (2) => (1)

page 103 in palace

f : I -> R, I is an interval on the real line

(1) for every x,y, every a,b>=0 such that a+b=1

f(ax+by)<=af(x) + bf(y)

(2) for every x<y, for every x<c<y

(f(c) - f(x)) / (c-x) <= (f(y) - f(c)) / (y-c)

(1) for every x,y, every a,b>=0 such that a+b=1

f(ax+by)<=af(x) + bf(y)

(2) for every x<y, for every x<c<y

(f(c) - f(x)) / (c-x) <= (f(y) - f(c)) / (y-c)

Hint:

a,b>=0, a+b=1,

c = ax + by,

x<=c<=y,

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

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

a,b>=0, a+b=1,

c = ax + by,

x<=c<=y,

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

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

(1) J is an open and connected subset of |R

(2) f:J->|R is a convex function

Does f have to be continuous?

(2) f:J->|R is a convex function

Does f have to be continuous?

YES.

page 115 in palace

page 115 in palace

f:A->B; g:B->R; A,B are intervals on the real line.

f,g are convex and g in increasing

Does g(f(x)) have to be convex?

f,g are convex and g in increasing

Does g(f(x)) have to be convex?

YES.

Let I be an interval on the real line. Let f : I -> |R.

(1) /\x,y:-I f(x/2 + y/2) <= f(x)/2 + f(y)/2

(2) f is continuous

Does f have to be convex?

(1) /\x,y:-I f(x/2 + y/2) <= f(x)/2 + f(y)/2

(2) f is continuous

Does f have to be convex?

Yes.

Show that for every natural n, for all natural p,q such that p+q=2^n,

f(p/2^n * x + q/2^n * y) <= p/2^n * f(x) + q/2^n * f(y).

Then show that an appropriate set is dense and then use continuity to conclude the thesis.

Show that for every natural n, for all natural p,q such that p+q=2^n,

f(p/2^n * x + q/2^n * y) <= p/2^n * f(x) + q/2^n * f(y).

Then show that an appropriate set is dense and then use continuity to conclude the thesis.