By $X$ we denote some pre-Hilbert space (it means that there is the inner product $\langle \cdot, \cdot \rangle$ but the corresponding normed space is not necessarily Banach). It can be proved that in the case of non-Hilbert spaces there exists a closed subspace $H_0 \subset X$ such that $H_0 \bigoplus H_0^\perp \ne X$.
So, let $X$ be the space of continuous functions $C([-1,1],\mathbb{C})$ endowed with the inner product $\langle f, g \rangle =\int_{-1}^{1}f(t)\overline{g(t)}dt$. Consider the subspace $$H_0=\left\{f\in X: \int_{-1}^0f(t)dt=\int_{0}^{-1}f(t)dt\right\}.$$
I'd like to understand whether $H_0\bigoplus H_0^{\perp} = X$. First of all, could you give me some hints to prove that $H_0 \bigoplus H_0^{\perp} = X$ indeed? (if my memory does not fail me, $X$ is not Banach, hence we cannot use the theorem about Hilbert spaces) However, I'd like to understand whether it is possible to disprove that without an example, using only some properties of $H_0$ itself?