[SOLVED] Best approximation/least squares/inner product

**arbolis** · June 18th 2008, 12:33 PM

Let $f^*=\sum_{j=0}^n c_j^* \phi_j$ be the best approximation with respect to f for the least squares, where { $\phi_0,\phi_1,...,\phi_n$ } is a linear independent set. Show that $\parallel f-f^* \parallel={\parallel f \parallel}^2-{\parallel f^* \parallel}^2$ where the norm is the one induced by the inner product.
My work : I've been up to $\parallel f-f^* \parallel={\parallel f \parallel}^2-{\parallel f^* \parallel}^2-2<f,f^*>$ . So if I show that $<f,f^*>$ =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?

**kalagota** · June 19th 2008, 05:44 AM

Originally Posted by arbolis

Let $f^*=\sum_{j=0}^n c_j^* \phi_j$ be the best approximation with respect to f for the least squares, where { $\phi_0,\phi_1,...,\phi_n$ } is a linear independent set. Show that $\parallel f-f^* \parallel={\parallel f \parallel}^2-{\parallel f^* \parallel}^2$ where the norm is the one induced by the inner product.
My work : I've been up to $\parallel f-f^* \parallel={\parallel f \parallel}^2-{\parallel f^* \parallel}^2-2<f,f^*>$ . So if I show that $<f,f^*>$ =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 $\parallel f-f^* \parallel \leq {\parallel f \parallel}^2-{\parallel f^* \parallel}^2$ and $\parallel f-f^* \parallel \geq {\parallel f \parallel}^2-{\parallel f^* \parallel}^2$

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

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