Step in the Stein's proof of Plancherel Theorem I am reading Stein's Fourier Analysis, and I am confused by the proof of Plancherel's theorem. 

If $f \in \mathcal S(\mathbb R)$, then $||f|| = ||\hat{f}||$. 

The norm here is defined in the $L^2$ sense. The proof goes like this.

If $f \in \mathcal S(\mathbb R)$, define $g(x) = \overline{f(-x)}$. 
  Then $\hat g(\xi) = \overline{\hat f(\xi)}$.
  Let $h = f*g$.
  Then $\hat h = \hat f \hat g = |\hat f|^2$, and $\displaystyle h(0) = \int f(-t)g(t) dt = \int f(-t)\overline{f(-t)} dt = \int|f|^2$. 
  Apply Fourier inversion formula, and get $\displaystyle h(0) = \int \hat h(\xi) d\xi$. Thus $\displaystyle\int |\hat f|^2 = \int \hat h = h(0) = \int |f|^2$.

I understand every step in this proof except for the setence $\hat g(\xi) = \overline{\hat f(\xi)}$. 
I try to expand this thing, and get
$\displaystyle \hat g(\xi) = \int g(x)e^{-2i\pi x\xi} dx = \int \overline{f(-x)}e^{-2i\pi x\xi} dx = -\int \overline{f(u)}e^{2i\pi u\xi}du = -\overline{\hat f(\xi)}$,
which would make everything after this off by a minus sign. Could someone tell me what went wrong with my argument here?

Your equality
$$\int \overline{f(-x)}e^{-2i\pi x\xi} dx = -\int \overline{f(u)}e^{2i\pi u\xi}du$$
is wrong because you simply forgot to "swap" the bounds:
$$\int_{-\infty}^{\infty} \overline{f(-x)}e^{-2i\pi x\xi} dx = 
-\int_{+\infty}^{-\infty} \overline{f(u)}e^{2i\pi u\xi}du = 
\int_{-\infty}^{+\infty} \overline{f(u)}e^{2i\pi u\xi}du $$
