I'm new to functional analysis. I do not know how to solve the below problem. I know that for g to be the best approximation of f, it has to satisfy the condition that (f−g)⊥G. But I could not apply it in this case.
Consider C[-1,1] (space of continuous functions in [-1,1]) and G:={g $\epsilon$ C[-1,1] | g(x)=g(-x) for all x}. Assume that C[-1,1] is equipped with the standard L^2[-1,1] inner product. Find the best approximation of $x^{{3}}$ + $x^{{2}}$ + $e^{{-x}}$ in G.
Could you please help me with how to approach the solution?
Thank you.
Let $(Ff)(x) = f(-x)$. Note that $g \in G$ iff $Fg =g$.
Note that we can write $f$ in its odd & even parts with $f = {f-Ff \over 2} + {f+Ff \over 2} $.
Then, grinding through the computations we get \begin{eqnarray} \|f-h\|^2 &=& \| {f+Ff \over 2} + {f-Ff \over 2} -g \|^2 \\ &=& \| {f+Ff \over 2} -g + {f-Ff \over 2} \|^2 \\ &=& \| {f+Ff \over 2} -g \|^2 + \| {f-Ff \over 2} \|^2 + 2 \langle {f+Ff \over 2} -g , {f-Ff \over 2} \rangle \\ &=& \| {f+Ff \over 2} -g \|^2 + \| {f-Ff \over 2} \|^2 \end{eqnarray} Note that ${f+Ff \over 2} \in G$, hence we have $\min_{g \in G} \|f-g\| = \| {f-Ff \over 2} \|$ and the minimising $g$ is given by $g={f+Ff \over 2} $.
Hence the best approximation is $x \mapsto x^2 + \cosh x$.
Note:
A much shorter version would be use the theorem you mentioned and to note that $\langle {f-Ff \over 2} , g\rangle = 0$ for all $g \in G$ and that ${f-Ff \over 2} = f - {f+Ff \over 2}$, from which it follows that $g= {f+Ff \over 2}$ is the solution.