1
$\begingroup$

This is given as a sample exercise for our final exam in our Numerical Analysis class. However, nowhere in the course has the notion of approximant been defined or how to approach this kind of problem . What I found on the web doesn't really seem compatible with this problem.

Let $g^* \in G$ represent the best approximant for $f \in F$, such that $<f-g,g^*> = 0 , \forall g \in G $

Prove that $g^*$ is unique.

Can anyone point the way to a solution ? If you don't have the time, I would also appreciate some references.

$\endgroup$
2
  • $\begingroup$ Are not $g$ and $g*$ reversed? It does not make sense $<f, g^*> = <g, g^*>$. The left is fixed, the right may be arbitrary. If the best approximant is defined as $<f, g> = <g^*, g>$ then $g^*$ is an orthogonal projection of $f$ on $G$. If there are two projections $g^*$ and $g^{**}$ their difference should be orthogonal to every $g$, thus zero, since $g^{**} - g^* \in G$. $\endgroup$
    – uranix
    May 31, 2015 at 22:17
  • $\begingroup$ They are definitely reversed. Must be a typo on OP's part $\endgroup$
    – Victor
    May 31, 2015 at 22:23

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .