Closed subspace of a functional space and its orthogonal complement Let's consider the following subspace of $C[-1, 1]$ with the following inner product $(f, g) = \int_{-1}^{1}{f(t) \overline{g(t)} dt}$: $$H_{0} = \{ f \in H | \int_{-1}^{0}{f(t) dt} = \int_{0}^{1}{f(t) dt} \}$$.
First, i would like to prove that $H_{0}$ is a closed subspace of $H$. This can be done, relying on the following theorem:

A linear functional on a normed space is bounded if and only if its kernel is closed.

For example, if a take $L(\varphi) = \int_{-1}^{0} {\varphi(t) dt} - \int_{0}^{1}{\varphi(t) dt}$, then $Ker(L)=H_{0}$.
How to show that $L$ is continous (or, equivalently, bounded operator)?
Moreover, i would like to establish, whether $H= H_{0} \oplus H_{0}^{\perp}$. If the $C[-1, 1]$ was complete in $||f||=\sqrt{(f, f)}$ it could be proven directly, since any  complete normed space can be decomposed into the direct sum of the closed subspace and its orthogonal complement. But this is not the case, as $C[-1, 1]$ is not complete in the given norm. Are there any hints that might help?
Any sort of help would be much appreciated.
 A: Often (not always) it's simpler to show the continuity of a linear functional on a normed space by explicitly giving a bound on its norm rather than showing that its kernel is closed. Here, we have
$$\lvert L(\varphi)\rvert = \biggl\lvert \int_{-1}^0 \varphi(t)\,dt - \int_0^1 \varphi(t)\,dt\biggr\rvert \leqslant \int_{-1}^1 \lvert \varphi(t)\rvert\,dt \leqslant \sqrt{2}\sqrt{\int_{-1}^1 \lvert \varphi(t)\rvert^2\,dt} = \sqrt{2}\lVert\varphi\rVert,$$
where the first inequality follows from the monotonicity of the integral, and the second by the Cauchy-Bun'akovskij-Schwarz inequality. Since we don't need it, I mention without proof that this bound is sharp, i.e. $\lVert L\rVert = \sqrt{2}$.
From the continuity, it follows that $H_0 = \ker L$ is closed.
Now, if we look at the completion of $C([-1,1])$ under the given norm, that is, $L^2([-1,1])$, we see that $H_0 = (\operatorname{span} \{\sigma\})^{\perp_{L^2}} \cap C([-1,1])$, where
$$\sigma(t) = \begin{cases}\; 1 &, t > 0 \\ \;0 &, t = 0 \\ -1 &, t < 0.\end{cases}$$
The fact that $\sigma$ has no continuous representative hints at the possibility that we may not have an orthogonal decomposition $C([-1,1]) = H_0 \oplus H_0^{\perp_C}$.
If we look at
$$H_1 = (\operatorname{span} \{\sigma\})^{\perp_{L^2}} = \left\{ [f] \in L^2([-1,1]) : \int_{-1}^0 f(t)\,dt = \int_0^1 f(t)\,dt\right\},$$
we know that $H_1^{\perp_{L^2}} = \operatorname{span} \{\sigma\}$, so $H_1^{\perp_{L^2}} \cap C([-1,1]) = \{0\}$.
Hence, if we can show that $H_0$ is dense in $H_1$, it follows that $H_0^{\perp_{L^2}} = H_1^{\perp_{L^2}}$ and hence
$$H_0^{\perp_C} = H_0^{\perp_{L^2}} \cap C([-1,1]) = \{0\},$$
which implies $C([-1,1]) \neq H_0 \oplus H_0^{\perp_C} = H_0$ (since e.g. nonzero constant functions are in $C([-1,1]) \setminus H_0$).
Indeed, $H_0$ is dense in $H_1$. To show that, use that the space of continuous functions with $f(0) = 0$ is dense in $L^2([-1,1])$, and approximate on the intervals $[-1,0]$ and $[0,1]$ separately. For the moment, I leave the details to you.
A: Hint: use the Cauchy-Schwarz inequality of $L^2([-1,1])$ to show that $L$ is bounded
$$\mid L(f)\mid\leq \int_{-1}^1\mid f(t)\mid dt= \langle\lvert f(t)\rvert,1\rangle_{L^2}$$
