# stuck on problem on completeness of subspace of C[0,1]

I am really stuck in the following problem:

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

I saw the following answer on stackexchange:

"Consider function $z(t)=t^2$ and the linear functional $$f:(C([0,1],\Vert\cdot\Vert_2)\to\mathbb{C}:x\mapsto \langle z,x\rangle$$ Obviously $f$ if continuous, hence its kernel is closed subspace of codimension $1$. Since $f(z)\neq 0$, then $X=\ker f\oplus\mathbb{C}z$. Note $z\in X_0^\perp$, and $X\oplus X_0^\perp\subset X$. Hence $X=X_0\oplus X_0^\perp$, and $X_0^\perp=X/\ker f=\mathbb{C}$. Thus $X_0^\perp$ is complete. If $X_0$ would be complete, then $X$ would be also complete because as direct sum of complete spaces. But $X$ is not complete, so does $X_0$."

The answer works fine for me but I do not understand the following aspect: We know that C[0,1] under given inner product is not a hilbert space (and $X_0$ is a closed subspace).Then,how can one say X=$X_0 +X_0^\bot$(since,this holds if X is a hilbert space and $X_0$ is a closed subspace of X).How has he expressed X as the direct sum.

I am really sorry if this question is not up to standard of stackexchange community.

any help would be appreciated...

• It can also be the case if $X$ is just a pre-Hilbert space and $X_0$ is a Banach subspace. – user384138 Jan 25 '17 at 9:14
• @Open Ball:could you please elaborate... – Abhishek Shrivastava Jan 25 '17 at 9:21

As pointed out by @Open Ball, we can relax some requirement for the orthogonal decomposition.

Indeed, if $W$ is a subspace of an inner product space $V$ and there exists an orthogonal projection $P$ from $V$ onto $W$, then the mapping

$$v \mapsto (Pv, v - Pv)$$

provides an isometry between $V$ and $W \oplus W^{\perp}$. A sufficient condition for the existence of such $P$ is that $W$ is complete w.r.t. the topology induced by the inner product.

A proof for this general claim is provided below, though this is easy to prove when $W$ is finite-dimensional. For instance, if $\dim W = n \in \Bbb{N}$ and $\{w_1, \cdots, w_n\}$ is an orthonormal basis of $W$, then

$$Pv = \sum_{k=1}^{n} \langle w_k, v \rangle w_k$$

is an orthogonal projection onto $W$. Now we can apply this fact to $V = X$ and $W = \Bbb{C}z$ with $z(t) = t^2$.

Lemma. Let $(V, \langle \cdot, \cdot \rangle)$ be a (real or complex) inner product space equipped with the topology induced by the inner product. If $W$ is a complete subspace of $V$, then there exists an orthogonal projection $P$ from $V$ onto $W$.

(In case of complex inner product, we adopt the convention that $\langle \cdot, \cdot \rangle$ is conjugate-linear in the first argument and linear in the second argument.)

Proof. Define the orthogonal projection $P : V \to W$ such that $Pv$ is the unique element in $W$ that minimizes the distance to $v$, or succinctly,

$$Pv := \underset{w \in W}{\arg\min} \| v - w \|.$$

Here, $\| \cdot \| := |\langle \cdot, \cdot\rangle|^{1/2}$ is the norm induced by the inner product. It is a routine job to prove that $P$ is well-defined:

• Existence. Let $\ell = \inf\{ \| v - w \| : w \in W \}$. Then for any $w, w' \in W$, the parallelogram law tells us that

\begin{align*} \|w - w'\|^2 &= 2\|w - v\|^2 + 2\|w' - v\|^2 - 4 \left\| \frac{w + w'}{2} - v \right\|^2 \\ &\leq 2\|w - v\|^2 + 2\|w' - v\|^2 - 4\ell^2. \tag{*} \end{align*}

Now for each $n$, we can find an element $w_n \in W$ such that $\|v - w_n\|^2 < \ell^2 + 2^{-n}$. Then by $\text{(*)}$,

$$\|w_n - w_{n+1}\|^2 \leq 2(\ell^2 + 2^{-n}) + 2(\ell^2 + 2^{-n-1}) - 4\ell^2 = 3\cdot 2^{-n}.$$

Thus $(w_n)$ is a Cauchy sequence. Since $W$ is complete, $(w_n)$ converges to some $w \in W$. This $w$ satisfies $\|v - w\| = \ell$ and hence $w$ is a distance-minimizer.

• Uniqueness. If $w, w' \in W$ minimizes the distance to $v$, then $\ell = \|v - w\| = \|v - w'\|$. By $\text{(*)}$,

$$\|w - w'\|^2 \leq 2\ell^2 + 2\ell^2 - 4\ell^2 = 0.$$

Therefore $w = w'$.

The mapping $P : V \to W$ is now well-defined and surjective. It remains to prove that $P$ is indeed an orthogonal projection to $W$. Again, establishing this claim is routine:

• Orthogonality. The proof follows from the usual variational principle. Let $v \in V$. Then for any $w \in W$, the quadratic function

$$t \mapsto \| v - Pv - tw \|^2 = \|v - Pv\|^2 - 2t\operatorname{Re}\langle v - Pv, w \rangle + t^2\|w\|^2$$

achieves the global minimum at $t = 0$. This implies that $\operatorname{Re}\langle v - Pv, w \rangle = 0$. Replacing $w$ by $iw$ in the complex inner product case, this implies that $\langle v - Pv, w \rangle = 0$. Since this is true for all $w \in W$, we have $v - Pv \in W^{\perp}$.

Conversely, assume that $w \in W$ satisfies $v - w \in W^{\perp}$. Then by the Pythagoras theorem

$$\ell^2 = \| v - Pv \|^2 = \| v - w \|^2 + \| w - Pv \|^2 \geq \ell^2 + \| w - Pv \|^2.$$

This shows that $w = Pv$. Thus $Pv$ is specified as the unique element $w \in W$ such that $v - w \in W^{\perp}$.

• Linearity. If $u, v \in V$ and $\alpha, \beta$ are scalars, then for any $w \in W$

$$\langle (\alpha u + \beta v) - (\alpha Pu + \beta Pv), w \rangle = \bar{\alpha} \langle u - Pu, w \rangle + \bar{\beta} \langle v - Pv, w \rangle = 0$$

and hence $\alpha Pu + \beta Pv = P(\alpha u + \beta v)$.