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.
 A: 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).
A: 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.
A: 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 $\int_0^1h(x)^2dx=1/2.$
