I am trying to prove a very simple estimate for extensions (Theorem 8.6) from the Brezis, i.e.,
$\|Pu\|_{L^{p}(\mathbb{R})} \leq 2 \|u\|_{L^{p}(I)}$
where $I = (0,\infty)$ and
$(Pu) (x)= \begin{bmatrix} u(x)\quad \text{for} \; x\geq 0, \\u(-x) \quad \text{for} \; x<0\end{bmatrix}$
Likewise, the estimate: if $u^* = (Pu) \in W^{1,p}(\mathbb{R})$ then $\|Pu\|_{W^{1,p}(\mathbb{R})} \leq 2 \|u\|_{W^{1,p}(I)}$.
If I use the definition, I end up with an equality while it is an inequality in the proof.
Using the definition,
$\|Pu\|_{L^{p}(\mathbb{R})}^p=\int_{-\infty}^{\infty}|Pu|^p dx = \int_{-\infty}^{0}|u(-x)|^pdx + \int_{0}^{\infty}|u(x)|^p dx $.
Change of variables and reversing the order of integration,
$\|Pu\|_{L^{p}(\mathbb{R})}^p = \int_{0}^{\infty}|u(y)|^p dy + \int_{0}^{\infty}|u(x)|^p dx ,$
Where I used $y=-x$ and reversed the order of integration in the first integral. Finally, I get
$\|Pu\|_{L^{p}(\mathbb{R})}^p = 2\int_{0}^{\infty}|u(x)|^p dx = 2\|u\|_{L^p(I)}^p.$
Can someone suggest where I am making the mistake, as it is trivial but I am kinda stuck..... Thanks in advance.