# Thread: [SOLVED] Best approximation/least squares/inner product

1. ## [SOLVED] Best approximation/least squares/inner product

Let $\displaystyle f^*=\sum_{j=0}^n c_j^* \phi_j$ be the best approximation with respect to f for the least squares, where {$\displaystyle \phi_0,\phi_1,...,\phi_n$} is a linear independent set. Show that $\displaystyle \parallel f-f^* \parallel={\parallel f \parallel}^2-{\parallel f^* \parallel}^2$ where the norm is the one induced by the inner product.
My work : I've been up to $\displaystyle \parallel f-f^* \parallel={\parallel f \parallel}^2-{\parallel f^* \parallel}^2-2<f,f^*>$. So if I show that $\displaystyle <f,f^*>$=0, then I'm done. I'm stuck on this. (I think I must use the fact that the set is linear independent... but how?) Can you help me?

2. Originally Posted by arbolis
Let $\displaystyle f^*=\sum_{j=0}^n c_j^* \phi_j$ be the best approximation with respect to f for the least squares, where {$\displaystyle \phi_0,\phi_1,...,\phi_n$} is a linear independent set. Show that $\displaystyle \parallel f-f^* \parallel={\parallel f \parallel}^2-{\parallel f^* \parallel}^2$ where the norm is the one induced by the inner product.
My work : I've been up to $\displaystyle \parallel f-f^* \parallel={\parallel f \parallel}^2-{\parallel f^* \parallel}^2-2<f,f^*>$. So if I show that $\displaystyle <f,f^*>$=0, then I'm done. I'm stuck on this. (I think I must use the fact that the set is linear independent... but how?) Can you help me?
i don't think if this would help..

my idea is to show that $\displaystyle \parallel f-f^* \parallel \leq {\parallel f \parallel}^2-{\parallel f^* \parallel}^2$ and $\displaystyle \parallel f-f^* \parallel \geq {\parallel f \parallel}^2-{\parallel f^* \parallel}^2$

you are already done with the first job, just do the second one..