# functions with same area

I have two real valued functions $f_1$ and $f_2$ such that

• $\int_0^Tf_1=\int_0^Tf_2=a_1$
• $\int_0^Tf_1^2=\int_0^Tf_2^2=a_2$
• $\forall \\ t, f_1(t),f_2(t)\in[0,1]$

Now,I want to construct a family of functions using $f_1$ and $f_2$ which also have all the above mentioned properties. For example $cf_1+(1-c)f_2$ have the first and third property and $\sqrt{cf_1^2+(1-c)f_2^2}$ have the second and third property.

But I want the functions to satisfy all three. Is there any way to combine the two functions in the required manner. Necessary smoothness conditions could be assumed.

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Abstractly you're intersecting a sphere in a real Hilbert space with a hyperplane. The hyperplane is a closed affine-linear subspace, by Cauchy-Schwarz. So the geometry isn't too bad. You seem to want some parametric information on the intersection. "Combining the functions" to you seems to mean working in the algebra they generate. Which is in the territory of the Stone-Weierstrass theorem, though that's for the complex (uniform) algebra, applying to continuous functions. There must clues in the functional analysis, but your formulation suggests you want a simple answer. –  Charles Matthews Apr 14 '12 at 14:03
@Charles: Yes I am interested in the parametric information and I just want a simple answer. Looking at functions as points in $L_2$3 does make it easy to visualize and may be also prove some existence results but is not helping in writing a parametrized version. @Mark: Getting an $F$ like you suggested in your first comment would be ideal. And I think there is one plane and one surface of a sphere (after doing a simple normalization), and we are interested in their intersection. After getting this intuition, geometry is no longer helpful to me. I really hope you could get something –  ConfuseD Apr 14 '12 at 15:40
@Igor: According to a well-known reference site, "A Hilbert space H is a real or complex inner product space that is also a complete metric space with respect to the distance function induced by the inner product." –  Charles Matthews Apr 15 '12 at 10:59
Be careful, however (especially with Fourier): you can easily go outside the range $[0,1]$ this way! In general there is no simple recipe. Imagine that you have $\alpha_1=\alpha_2=m$. In this case, the only solutions are characteristic functions of sets of measure $m$. They can be moved one to another, no question, but you need more good luck than there is out there doing it with an explicit algebraic formula! –  fedja Apr 15 '12 at 11:49
@fedja: next you will tell me 57 is not prime, what's the world coming to... –  Igor Rivin Apr 16 '12 at 4:31

Fourier series. If I'm not being slow, writing the functions as Fourier series of period T allows one to remove the first condition, as the zeroth Fourier coefficient being fixed. Then you want to navigate with the sum of the absolute squares being fixed. Basically this is paths on a sphere with one dimension taken out. There's a rotation that would do this, isn't there?

'''Edit''': The question as posed afresh involves a uniform norm bound also (at most 0.5 from the constant function 0.5, to put in one way).

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You don't actually need Fourier series. Any two functions lie in a two dimensional linear space (the one containing the two of them and zero). In two dimensions, the problem cannot be solved (since the intersection of a circle and a line usually not connected). However, the intersection of a sphere and a plane is, so manufacture some third function $f_3$ with your property which is not a linear combination of the two functions you had started with. then your question reduces to a simple question in three-dimensional geometry.

EDIT The only property $f_3$ should have is being linearly independent of $f_1, f_2$ that gives you the three dimensional space to work on. For this, maybe Fourier series is the simplest way: just pick the three-term approximations to $f_1, f_2$ and pick a combination of the three harmonics linearly independent from these (by taking the vector product of the coefficient vectors, e.g.).

Further edit The answer above is for the first two conditions. Once you introduce a sup norm condition, things become very difficult. Indeed, as pointed out by @Greg Kuperberg and @Bill Johnson in their answers to this question, every finite dimensional Banach is a finite dimensional slice of $L^\infty,$ so the space you need to check connected is the intersection of the unit sphere with some arbitrary centrally symmetric convex set. That makes the problem much harder. To summarize: first two conditions: easy, last condition: hard.

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@Igor: Yes Series would work but is kind of an overwork here. Also, finding $f_3$ is, in essence, the problem here. –  ConfuseD Apr 15 '12 at 7:47
@ConfuseD: See the edit -- I am not sure trying to keep out the Fourier series completely is wise, although there might be some series-free method of producing an $f_3$ such that $\int f_1 f_3 d x = \int f_2 f_3 d x = 0.$ –  Igor Rivin Apr 15 '12 at 12:27
Igor, the life is not this simple. See my remark to the main post... –  fedja Apr 15 '12 at 17:28
@fedja: I understand now, it is the sup norm is the real problem -- th restriction of it to the subspace could be a (presumably arbitrary) convex set, which is why algebraic solutions are not likely to work in general.... –  Igor Rivin Apr 16 '12 at 23:10

Given just $f_1$ you can do a cyclic shift $F(x,c)=f_1(x+c)$ for $x \lt T-c$ but $F(x,c)=f_1(x+c-T)$ for $x \gt T-c$. But you want to use both.

A scrunched up version could work. For convenience I am going to assume $T=1.$ Let $F(x,0)=f_1(x),F(x,1)=f_2(x)$ and $F(x,c)=f_1(\frac{x}{c})$ for $0 \le x \le c$ but $F(x,c)=f_2(\frac{x-c}{1-c})$ for $x \gt c.$

Your two examples of the type of thing you really want start with a function $g(u,v,c)$ of two variables $u,v$ and a parameter $c$ with the added property $g(u,v,c)=g(v,u,1-c)$ and then set $F(x,c)=g(f_1(x),f_2(x),c).$ I don't think all that is possible at the same time. Let $f_2(x)=1-f_1(x)$ where $f_1(x)=0$ or $1$ according as $0 \le x \le 1/2$ or $1/2 \lt x \le 1$ Then $h(x)=F(x,1/2)$ would be a constant function on $[0,1]$ and hence unable to have $\int_0^1h(x)dx=1/2$ and also $\int_0^1h(x)^2dx=1/2.$

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