In the vector space $C(1,3)$, endowed with the internal product $$\langle f, g \rangle = \int_ {1}^{3} f (x) g (x) dx $$ for all $f, g \in C (1,3) $. For $f (x) = \dfrac{1}{x} $ with $x \in (1,3) $, show that the constant polynomial $g $ closest to $f $ is $g (x) = \dfrac{\ln (3)}{2} $. Calculate $|| f-g || ^ {2} $
What do you mean with a closer function?
We have $$ ||f-g||^2 = \int_1^3 (f-g)^2(x)dx = \int_1^3 (1/x-g)^2dx = 2 g^2 - g \log(9) + 2/3 =p(g) $$ The parabola $p(g)$ has a minimum at $g^*=\log(3)/2$ and $||f-g||^2$ is precisely $$p(g^*)=\frac{4 - 3 \log^2 3}{6}$$