Targil 3 - functions.
1
1
1
, where a1, … an are

 ... 
x  a1 x  a2
x  an
some real constants. Compute the total length of f 1  a, b .
1. Consider a function f  x   x 
(Here  a, b is an interval between a and b, f 1  set  denotes the inverse image of
that set under f, that is all points that are sent by f to that set, and total length of
several intervals is the sum of their lengths).
Answer. b – a, in other words this transformation preserves length!
Solution. All summands are locally increasing, so f is increasing where it is
defined. It is defined at all points except a1, a2, … , an. When x approaches one of
those points from below, f tends to infinity, and when x approaches one of those
points from above, f goes to minus infinity. Hence if we split the real axis into n+1
open intervals  , a1  ,  a1, a2  ,  a2 , a3  ,...,  an ,   , on each interval f is monotone
increasing, and values cover all the real axis. So, for each real y we shall have n+1
inverse images, x0, x1, …, xn, one for each of the above intervals, and they increase
monotonically when y increases.
So, assume that inverse images of a are u0, u1, u2, … , un and inverse images of b
are v0, v1, v2, … , vn. Then inverse image of [a,b] consists of n+1 intervals [ui, vi],
for 0 ≤ i ≤ n. Its total length can be computed as
n
n
n
v  u   v  u .
i
i 0
i
i 0
i
i 0
i
1
1
1
.

 ... 
x  a1 x  a2
x  an
This equation can be turned into polynomial by multiplying by all denominators
and moving everything to the right hand side (RHS). The highest term will be xn+1
and the next will be – (a+a1+a2+…+an)xn, and all the terms that come from the
fractions will be of lower degree. So, by Vieta theorem, the sum of all n+1 roots is
Now, all ui are zeroes of an equation a  x 
n
u
i 0
i
n
 a   ai . Similarly,
i 1
n
v
i 0
total length of inverse image =
i
n
 b   ai . Back to the original question: the
i 1
n
n
n
i 0
i 0
i 0
  vi  ui    vi   ui  b  a .


2. Let f  x   2 x 1  x  , for xR . Define f n  x   f ... f  f  x  ... , where f is
applied n times.
1
(a) Compute lim  f n  x  dx .
n 
0
1
 f  x  dx for all natural n.
(b) Compute
n
0
x 1  x  
Solution. (a) On [0,1],
hence f  x   2 x 1  x   2  12   12 .
x  1  x  1

2
2
2
If 0 < x < ½, then f  x   x   2 1  x   x .
Consider the sequence xi  fi  x  . Then for i > 0, 0  xi  1 .
Hence for i > 0, it is monotonically increasing, xi  xi 1 , and bounded by ½ from
above, so it has limit ≤ ½ , x*  lim xi , and x*  f  x*  , so x*  1 2 .
i 
On the 0 < x < ½, f is monotonic increasing, and so is fn.
So, for every  for n great enough, we shall have 12    f n    12 .
So for all   x  12 , we shall have 12    f n    12 .
Since f and fn are symmetric w. r. t. ½, same inequality hold for 1  x  1   .
2
On the other subintervals, 0,   and 1   ,1 , we have 0  f n     1 . So,
2

1
1
2
   1  2    f n  x  dx 
1
2
0
Therefore, for large n, integral is close to ½.
(b) Denote xi  fi  x  , x0  x, yi  1 2  xi .
Then we can express yi+1 as a function of yi:
 1  xi 
2
yi 1  2  xi 1  2  2 xi 1  xi   1  4 xi  4 x   2 
  2 yi .
 2 
2
1
Therefore: 2 yi 1   2 yi 
2
i
1
2
1
2

2 
So f n  x   xn  12  yn  12  12  2  12  x  
1

0
1
f n  x  dx   12  12  2  12  x  
0

2
2 2
 2n 
, and hence 2 yn    2 y0  ...    2 y0  .


2 
n
n
.
1 1  2  x 
dx   
2 4
2n  1
1
2
 2 1
n
1

0
1
1

2 2  2n  1
Remark. In the above problem, the approach of (b) is shorter than that of (a) and
gives more results, so one could say that (a) is misleading and write (b) right away.
However the idea of (a) is more simple and it gives some intuition that the final
answer should be around ½ . To eliminate computational errors in the second term,
if any, substitute n = 1.
3. Prove that there is no function f : R  R such that f  0   0 and
f  x  y   f  x   y  f  f  x   for all x, y  R .
Solution. Suppose that there exists a function satisfying that inequality. If
f  f  x    0 for all x, then f is decreasing function, because
f  x  y   f  x   y  f  f  x    f  x  for negative y.
Since f  0  0  f  f  x   and the function is decreasing, it means 0  f  x  for
all x, but it is not true, since 0  f  f  x   . That is a contradiction.
So, for a certain value of z, we have f  f  z    0 .
Then f  z  y   f  z   y  f  f  z   so f is bounded from below by growing
linear function, therefore lim f  x    and hence lim f  f  x     .
x 
x
Take x  0 such that f  x   0 and f  f  x    1 .
Take y  0 such that y 
x 1
and f  f  x  y  1   0 and f  x  y   0 .
f  f  x   1
Then
f  x  y   f  x   y  f  f  x   f  x   y  f  f  x  


 y  f  f  x   1  y  x  y  1
and hence
f  f  x  y    f x  y  1   f  x  y    x  y  1  


 f  x  y  1   f  x  y    x  y  1  f  f  x  y  1  
 f  x  y  1  f  x  y   f  f  x  y   
 f  f  x  y 
A contradiction, QED.
4. Prove that for every continuous function f :[0,1  0,1  R ,
1 1
  f  x, y  
0 0
2
2
2
2
1 1
1 1
 11



dxdy    f  x, y  dxdy      f  x, y  dx  dy     f  x, y  dy  dx
00
00
 00


First solution. Actually, that is a question about geometry of linear spaces and not
about analysis. The continuous functions 0,1  0,1  R form a linear space.
1 1
They can be added and multiplied by numbers. There is a norm
  f  x, y  
2
dxdy
0 0
which make it similar in many ways to Euclidean space. It comes from a scalar
1 1
product f , g   f  x, y  g  x, y  dxdy .
0 0
Remark. For complex valued functions, the scalar product should rather be
1 1
defined as f , g   f  x, y  g  x, y dxdy .
0 0
1
For each function f we can construct a function A  f    f  t , y  dt .
0
This is function of y only, but we can say it is a function of x and y, just it doesn’t
1
really depend on x. In the same way we can define B  f    f  x, t  dt .
0
Objects like A and B are usually called operators: they take one element of a vector
space and make another.
Definition. An operator P is called a projection, if P2 = P. It is called an
orthogonal projection, if P(x) is orthogonal to x – P(x) for every x.
Here, by orthogonality of vectors we mean zero scalar product.
Remark. These notions are standard.
Exercise. A and B are orthogonal projections.
It is easy to see that AB  f  is constant, and that constant is equal to the integral
over the whole unit square, and that AB  f   BA  f  .
After setting up the correct framework, the solution is a game of few lines with
scalar product:
Denote A f   a, B  f   b, A B  f    B  A f    c .
The claim is f , f  c, c  a, a  b, b .
We can write f  a  u , where a, u  0 .
Similarly, f  b  v , where b, v  0 .
Also, B  a   c , so a  c  w , where c, w  0 .
b  B  f   B  a  u   B(a )  B  u   c  B  u  .
But c  A  b  is an orthogonal projection of b, so c, B  u   0 .
Orthogonal projection can only make a vector shorter, (why?), Therefore
f , f  c, c  a  u, a  u  c, c  a, a  u, u  c, c 
 a , a  B  u  , B  u   c , c  a , a  B  u   c , B  u   c  a , a  b, b
Second solution. This problem comes from Putnam, and the solution there was
with two-dimensional Fourier series. Actually, Fourier series is just decomposing
into a basis where those projections will look nice:
2 i nx  my 
If f  x, y    an,me 
, then
nZ mZ
1 1
  f  x, y  
2
2
dxdy   an ,m
2
,
nZ mZ
0 0
2
1

2
f
x
,
y
dx
dy

a
,





0,m
0  0
mZ

The inequality follows directly.
1
1 1

2
  f  x, y  dxdy   a0,0 ,
00

2
1

2
f
x
,
y
dy
dx

a





n ,0
0  0
nZ

1
Third solution. (Dan Carmon) A2  B2  2 AB .
In particular, for each quadruple x, y, z, w:
 f  x, y   f  z , w     f  x , w   f  z , y  
2
2
 2  f  x, y   f  z , w    f  x , w   f  z , y   .
Now divide by 4 and integrate over the cube  0,1 .
That part is left as an exercise to the reader.
What are you waiting for?
QED.
Remark. Equality takes place when A  B that is
f  x, y   f  z, w  f  x, w  f  z, y  , which is the same as f  x, y   g  x   h  y 
4
5*. Can a minimal value of a polynomial with rational coefficients be
2?
By minimal value here we mean the value at a point of global minimum.
Solution. Quoting from http://taharut.org/students/students2_solved.pdf :
Let us try p '  x   k  x 2  2   x  a   x 3  ax 2  2 x  2a .
 x 4 ax3

p x  k  
 x 2  2ax   c .
3
 4

The extremal points are  2 , a, so when we substitute them into p(x) we have
good chances to get something with  2 . Of course, if k > 0, then the middle
extremum is a maximum, and the other two are minima.
 2a 2

2

p  2  k 1
 2  2a 2   c  k  1  2a 2   c
3
3





3
. Then that will be the middle extremum.
4
The local minima are at  2 , and the global minimal value is the least between
Choose a 

 


 

p  2  k 1  2  c , which is p  2  k 1  2  c . Take k = c = 1
End of quote.
Therefore,
3
2 will be minimum for x  .
4
So it will be a global minimum of p  34  x 2  , which is a polynomial of degree 8.
Question**. What is the lowest possible degree of the solution to that question? It
seems there is no solution of degree 4. Is there any solution of degree 6?