problem on hilbert spaces

Question

Let $X=C[0,1]$ with the inner product $\langle x,y\rangle=\int_0^1 x(t)\overline y(t)\,dt$ $\forall$ $x(t),y(t)\in C[0,1]$

$X_0 =\{x(t) \in X :\int_0^1 t^2x(t)\,dt=0\}$and $X_0^\bot$ be the orthogonal complement of $X_0$.

(A) which of the following is correct:

(1)both $X_0$ and $X_0^\bot$ are complete

(2) neither $X_0$ nor $X_0^\bot$ is complete

(3)$X_0$ is complete but $X_0^\bot$ is not complete

(4) $X_0^\bot$ is complete but $X_0$ is not complete

(B) let $y(t)=t^3$, $t\in [0,1]$ and $x_0\in X_0^\bot$ be the approximate of $y$. Then $x_0(t)$,t$\in$[0,1] is

(1)$\frac{4t^2}{5}$

(2)$\frac{5t^2}{6}$

(3)$\frac{6t^2}{7}$

(4)$\frac{7t^2}{8}$

i am really stuck in these problems and can't proceed after a few steps.i would be really thankful if someone can give satisfactory explanations to what should the answers be...

thanks in advance....

Answer 1

$X=C[0,1]$ is not complete with respect to the norm $\|f\|=(f,f)^{1/2}=\int_0^1|f(t)|^2dt$. The completion of $C[0,1]$ with respect to this norm is $L^2[0,1]$. Every $f \in X$ can be written as $$ f = \left[f-\frac{(f,t^2)}{(t^2,t^2)}t^2\right]+\frac{(f,t^2)}{(t^2,t^2)}t^2 $$ The vector in square brackets is orthogonal to $t^2$. So $X=X_0\oplus[\{t^2\}]$, where $[\{t^2\}]$ consists of all scalar multiples of $t^2$. The subspace $[\{t^2\}]$ is complete because every finite-dimensional subspace of an inner product space is complete. You can use the above decomposition to show that $f \in X_0^{\perp}$ iff $f \in [\{t^2\}]$. That is, $X_0^{\perp}=[\{t^2\}]$. If $X_0$ were complete, then the above decomposition would imply $X$ is complete, which it is not. So the answer is (4): $X_0^{\perp}=[\{t^2\}]$ is complete, but $X_0$ is not complete. For the second part, \begin{align} t^3 & = \left[t^3-\frac{(t^3,t^2)}{(t^2,t^2)}t^2\right]+\frac{(t^3,t^2)}{(t^2,t^2)}t^2 \\ & = \left[t^3-\frac{4}{5}t^2\right]+\frac{4}{5}t^2. \end{align}

Answer 2

A metric space is complete iff every Cauchy sequence converges. If we want to prove that $X_0$ is incomplete, we have the advantage that all of our spaces are subspaces of the larger Hilbert space $L^2[0,1].$ Find a sequence in $X_0$ that converges (in $L^2$) to a discontinuous function. The sequence is Cauchy (that property only depends on the metric, not on the embedding space) and its limit, necessarily unique, lies outside $X_0.$

For the orthogonal complement of $X_0$ note that $X_0$ itself is already defined as the orthogonal complement of the singleton $f(t)=t^2.$ Use this to prove that $X_0^\bot$ is one-dimensional, and therefore complete.

In (B) we are asked to approximate the function $g(t)=t^3$ by the best fitting multiple of $f(t)=t^2.$ First normalize $f$ to a unit vector (divide it by its own norm). Now compute the scalar product of $g$ with that unit vector. Put that scalar product in front of the unit vector and you obtain the orthogonal projection (i.e., the best approximation) of $g$ on $X_0^\bot.$