Orthornormal basis in L^([a,b])

[tkjacobsen]

Homework Statement

(e_n) is orthonormal basis for L^2([0,1]).

Want to show that (f_n) is basis for L^2([a,b]) when f_n(u) = (b-a)^{-1/2}e_n(\frac{u-a}{b-a})

Homework Equations

f_n(u) = (b-a)^{-1/2}e_n(\frac{u-a}{b-a})

The Attempt at a Solution

I did show that (f_n) is an orthonormal sequence. But how can I show that it is also a basis...

 

[HallsofIvy]

Since they are orthonormal they are necessarily independent so you only need to show that they span the space. Given any f in L²([a, b]) you need to show that f is equal to \sum a_n f_n. Use your "relevant equation" to rewrite that in terms of e_n and use the fact that {e_n} spans L²([0, 1]).

 

[tkjacobsen]

but then i get
\sum a_n f_n(u) = (b-a)^{-1/2}\sum a_n e_n( (u-a)/(b-a) )

Can I then just say that since e_n spans L^2([0,1]) then e_n( (u-a)/(b-a)) spans L^2([a,b]) so the above equals f for the right choice of a_n.

Or am I missing something

Thanks