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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

any help would be appreciated..

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    Hint: Write $X_0=\left\{x\in X\mid \left\langle v,x\right\rangle=0\right\}$ where $v(t)=t^2$. $X_0$ is thus by itself the orthogonal complement of the space spanned by $v$. – 2017-01-24

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Both are complete. To see this, it suffices to prove that both are closed in $C([0,1])$. Consider the functional

$$ I : C([0,1]) \to \Bbb{C}, \qquad I(x) = \int_{0}^{1} t^2 x(t) \, dt. $$

Then $I$ is a continuous linear functional on $C([0,1])$ and $X_0 = \ker(I)$. Thus $X_0$ is a closed subspace. Likewise, for each $x \in X_0$ we define

$$ L_x : C([0,1]) \to \Bbb{C}, \qquad L_x(y) = \int_{0}^{1} \bar{x}(t) y(t) \, dt. $$

Again, each $L_x$ is a continuous linear functional on $C([0,1])$ and $X_0^{\perp} = \cap_{x \in X_0} \ker(L_x)$. Since $X_0^{\perp}$ is an arbitrary intersection of closed subspaces, $X_0^{\perp}$ is also a closed subspace.

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    I understand the arguments. But to this be true, it is required to $X = C[0,1]$ be complete as well, isn't it ? – 2017-01-24
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    @IgorD. I assumed that $C([0, 1])$ is equipped with the topology induced by the supremum norm. Probably I should have mentioned this explicitly. If we equip $X$ with the $L^2$-topology, then of course $X$ is not complete. – 2017-01-24
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    I see! But since only the inner product is specified, I would assume that $X$ is equipped with the $L^2$ topology. If that is the case, what would be the answer ? – 2017-01-24
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    @IgorD. If we assume $L^2$-topology, then $X_0$ is not complete but $X_0^{\perp}$ is still complete. This is because $X_0^{\perp}$ is 1-dimensional. – 2017-01-24
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    @Sangchul Lee why is under $L^2$ norm ,$X_0$ not complete – 2017-01-24
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    @vikashtripathi, Set $v(t) = t^2$. Then we can check that $$ \overline{X_0}^{\| \cdot \|_{L^2}} \simeq \{ x \in L^2([0,1]) : \langle x, v \rangle = 0 \}. $$ That is, we can find a bunch of elements in the $L^2$-completion of $X_0$ that is not in $X_0$. A more elementary argument is as follows: Consider the projection operator $$ P(x) = x - \frac{\langle x, v\rangle}{\langle v, v \rangle} v. $$ If we choose $x \in L^2([0,1]) \setminus X$ and $(x_n) \subset X$ such that $x_n \to x$ in $L^2$, then $Px \notin X$ and $X_0 \ni Px_n \to Px$ in $L^2$. – 2017-01-24
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    @ Sangchul Lee thaks for the example – 2017-01-24
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    @SangchulLee, Why $X_0^\perp$ is one-dimensional ? – 2017-01-24
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    @IgorD. A simple argument is as follows: Let $x \in X_0^{\perp}$. Using the projection operator $P$ defined as above, we have $Px \in X_0$. So we have $\langle Px, x \rangle = 0$. This condition can be rewritten as $$ \langle x, x\rangle \langle v, v\rangle = | \langle x, v \rangle|^2. $$ That is, $x$ achieves the equality in the Cauchy-Schwarz inequality and hence is a constant multiple of $v$. – 2017-01-24
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    @SangchulLee Thank you! – 2017-01-24