Which of these two operators is an orthogonal projection? Which of these two operators on $L^2(\mathbb{R})$ is an orthogonal projection? $T(f(x)) = f(|x|)$ or $S(f(x))= f(-|x|)$? Should I use the definition (which I tried and failed at) or some other sufficient condition?
 A: It seems to me like the answer is neither. We see that the image of both $S$ and $T$ is equal to the subspace of even functions (for any even function $g$, we have $Tg=Sg=g$, and clearly, the image of $T$ and $S$ consists of even functions).
Let $Uf(x)=\frac{f(|x|)+f(-|x|)}{2}$ and note that for even $g$, we have
$$
\langle f,g\rangle=\int_{-\infty}^0 f(x)\overline{g}(x)\textrm{d}x+\int_0^{\infty} f(x)\overline{g}(x)\textrm{d}x=\int_0^{\infty} (f(x)+f(-x))\overline{g}(x)\textrm{d}x=\langle Uf,g\rangle,
$$
where the last equality follows from the same calculation since $Uf(x)+Uf(-x)=f(x)+f(-x)$. Again, it's clear that $U$ is a projection onto the space of even functions. From the above, it follows that $U$ is the orthogonal projection onto this space. However, $U$ is neither equal to  $T$ nor $S$.
Concretely, consider the $L^2$ functions  $f_1(x)=1_{x\leq -1}\frac{1}{x}$ and $f_2=1_{x\geq 1} \frac{1}{x}$ and note that $Tf_1=S f_2=0,$ but we see that $f_1+f_2$ is even and
$$
\langle f_1,f_1+f_2\rangle=\langle f_2,f_1+f_2\rangle=1\neq0,
$$ 
so neither $T$ nor $S$ is an orthogonal projection.
