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I am trying to solve this,,,need help....
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Since it has taken already too much time, I will give a shot.I've already explained that in my response #6. af(1)+ bf(0)+ cf(1)= 0 for all polynomials of order 2 or less. Take f to be x^{2}, x, and 1 and you get three equations to solve for a, b, and c.
So you are taking af(0)+ bf(1)+ cf(1)= 0? That's not what I had before but perfectly legal.Since it has taken already too much time, I will give a shot.
If f = x^2, then
f(0) = 0
f(1) = 1
f(1) = 1
=> a*0 + b*1 + c*1 = 0
You mean a*0+ b*(1)+ c*1= b+ c= 0, right?If f = x, then
f(0) = 0
f(1) = 1
f(1) = 1
=> a*0 + b*1 + c*1 = 0
No, b+ c= 0.If f = 1, then
f(0) = 1
f(1) = 1
f(1) = 1
=> a*1 + b*1 + c*1 = 0
So we have three equations
b + c = 0
b + c = 0
The three equations you got, b+ c= 0, b+ c= 0, a+ b+ c= 0, the first two being the same, would give b= c so that a+ b+ c= a c+ c= a= 0 but does NOT mean b and c must be 0. for example b= 1, c= 1, a= 0 satisfy b+ c= 1 1= 0, a+ b+ c= 0+ 1 1= 0. That would prove S* is NOT linearly independent.a + b + c = 0
Simplifying
b + c = 0 (1)
a + b + c = 0 (2)
Substituting the equation (1) to (2)
a = 0
By trial and error, we know that
b = 0
c = 0
Thus,
a + b + c = 0
This means that S* is linearly independent.

I leave the rest for the questioner.
So you are taking af(0)+ bf(1)+ cf(1)= 0? That's not what I had before but perfectly legal.
You mean a*0+ b*(1)+ c*1= b+ c= 0, right?
Reformatting what HallsOfIvy says[tex]f_{1}(x)[/tex] such that
[tex]\phi_{1}(f_{1})= f_{1}(1)= 1, [/tex]
[tex]\phi_{0}(f_{1})= f_{1}(0)= 0,[/tex]
[tex]\phi_1(f_{1})= f_{1}(1)= 0[/tex]
[tex]f_{0}(x)[/tex] such that
[tex]\phi_{1}(f_0)= f_{0}(1)= 0, [/tex] # I fixed 1 here for phi
[tex]\phi_{0}(f_0)= f_0(0)= 1,[/tex]
[tex]\phi_1(f_0)= f_0(1)= 0[/tex]
[tex]f_{1}(x)[/tex] such that
[tex]\phi_{1}(f_1)= f_{1}(1)= 0, [/tex] # I fixed 1 here for phi
[tex]\phi_{0}(f_1)= f_1(0)= 0,[/tex]
[tex]\phi_1(f_1)= f_1(1)= 1[/tex]
Since we are talking about polynomials of degree 2 or less, this reduces to
1) Find [itex]f_{1}(x)= ax^2+ bx+ c[/itex] satisfying
a) [itex]f_{1}(1)= a b+ c= 1[/itex] # I am not sure should this be 1
b) [itex]f_{1}(0)= c= 0[/itex]
c) [itex]f_{1}(1)= a+ b+ c= 0[/itex]
2) Find [itex]f_{0}(x)= ax^2+ bx+ c[/itex] satisfying
a) [itex]f_{0}(1)= a b+ c= 0[/itex]
b) [itex]f_{0}(0)= c= 1[/itex]
c) [itex]f_{0}(1)= a+ b+ c= 0[/itex]
3) Find [itex]f_{1}(x)= ax^2+ bx+ c[/itex] satisfying
a) [itex]f_{1}(1)= a b+ c= 0[/itex]
b) [itex]f_{1}(0)= c= 0[/itex]
c) [itex]f_{1}(1)= a+ b+ c= 1[/itex]