Define $K \subset L^2(\mathbb{R})$ (with $L^2(\mathbb{R})$ being the set of all square-integrable functions) as \begin{align} K = \{ f \in L^2(\mathbb{R}) \ | \ f(-x) = 2f(x) \ \text{for almost all} \ x \geq 0 \}.\end{align}
I have to prove and find an expression for $K^{\perp}$ and an explicit formula for the orthogonal projection $p_K(f)$ for all $f \in L^2(\mathbb{R})$. However, I have troubles finding an explicit orthonormal basis for $K$ (with this I mean a set of orthonormal functions such that $K$ is equal to the closure of the linear span of that set) as that's the only way I know to find a formula for $p_K(f)$ (and thus also find $K^\perp$). I haven't yet learned anything about Fourier series, so I don't think I will need that.
Suppose $g\in L^2$ is orthogonal to $K$. Then $$ \int_{-\infty}^{\infty}g(x)\overline{f(x)}dx =0,\;\; f\in K, $$ which may be written as $$ 0=\int_{-\infty}^{0}g(x)\overline{f(x)}dx+\int_{0}^{\infty}g(x)\overline{f(x)}dx \\ =\int_{0}^{\infty}g(-x)\overline{f(-x)}+g(x)\overline{f(x)}dx\\ =\int_{0}^{\infty}g(-x)2\overline{f(x)}+g(x)\overline{f(x)}dx \\ =\int_{0}^{\infty}(2g(-x)+g(x))\overline{f(x)}dx. $$ This holds for arbitrary $f\in L^2[0,\infty)$, which is equivalent to $g(x)=-\frac{1}{2}g(-\frac{x}{2})$ for a.e. $x \in (-\infty,0]$.