Two equivalent norms on a Sobolev space

Question

I asked a question (Norm in a Sobolev Space) a week ago. My professor added a new request to the exercise and, once again, I am stuck. I will sum up what I need to solve it: Let $W^{1,2}=W^{1,2}(\mathbb R)$ be the Sobolev space consisting of functions $f\in L^2(R)$ such that exists $f_w\in L^2(R)$ with the following condition: $$ \int_R f(x)\, \varphi' (x)\,dx=-\int_R f_w(x)\, \varphi (x)\,dx\qquad \forall \varphi \in \mathcal S(R) $$ where $S(R)$ is the Schwartz space.

I have to show that this two norms are equivalent on $W^{1,2}$:

(NB: $||\cdot||_1 $ and $||\cdot||_2$ are two equivalent norms if $\exists c$, $c' \in C$ which are positive such as $c||x||_2\le||x||_1\le c'||x||_2$.)

1. $$\|f\|_ {W^{1,2}}=\left(\|f\|_2^2+\|f_w\|_2^2\right)^{1/2} $$

2. $$\|f\|^2:= \|\widetilde f\|_2^2+ \|\xi \widetilde f\|_2^2, \hspace{0.4cm} \forall\, f\in W^{1,2}$$ where $\widetilde f$ is the Fourier transform of $f$ in $L^2(R)$ (using the variable $\xi$).

I was able to do the following:

$$\|\xi\widetilde{f}\|^2_2 = \int_R |\xi \widetilde f(\xi)\ |^2 d\xi= \int_R |\xi|^2 | \int_R f(x)e^{-2\pi i\xi x} dx |^2 d\xi= \spadesuit $$ Now I decided to use the fact that $f \in W^{1,2}$ : $$ \int_R f(x)\, \varphi' (x)\,dx=-\int_R f_w(x)\, \varphi (x)\,dx\qquad \forall \varphi \in \mathcal S(R) $$ Due to the fact that $e^{-2\pi i\xi x} \in S(R)$, I defined $\varphi '(x) = e^{-2\pi i\xi x}$.

This means that $\varphi (x) = -\frac{e^{-2\pi i\xi x}}{ 2\pi i \xi }$

$$ \spadesuit = \int_R |\xi|^2 |- \int_R f(x)\frac{e^{-2\pi i\xi x}}{-2\pi i \xi} dx |^2 d\xi= \frac{1}{4\pi^2} \int_R | \int_R f_w(x) e^{-2\pi i\xi x}dx|d\xi= \frac{1}{4\pi^2} \|\ \widetilde{f_w}\|\ ^2_2$$

$\Rightarrow \|\xi \widetilde{f}\|^2_2=\frac{1}{4\pi^2} \|\ \widetilde{f_w}\|\ ^2_2$

Another thing that I know is that on $L^2(R)$ is defined the following surjective isometry:

$ \| f\|_2=\| \widetilde{f} \|_2$ and using that I can state that: $ \|\xi f\|^2_2=\frac{1}{4\pi^2} \|\ f_w\|\ ^2_2$.

I assume that I can use this to show that the two norms are equivalent, but I don't know how. The thing that confuses me is that the norms are defined using the sums of two norms. Can someone help me? I have been trying for three hours so it might be obvious but I am really tired. Thank you!

Answer 1

Some issues with your approach:

1. $e^{-2\pi i\xi x}$ is not in $\mathcal{S}(\mathbb{R})$. It doesn't decay at infinity - it's not even integrable!

2. It will be easier to to this step by step: show that $\|f\|_{W^{1,2}} \leq C_1\|f\|$, and $\|f\| \leq C_2\|f\|_{W^{1,2}}$ for some positive constants $C_1$ and $C_2$.

Some advice:

1. The basic idea is sound: at some point in the argument, you do want to show that $\|f_w\|_2 = C\|\xi\tilde{f}\|_2$ for some constant $C$. But the easier way to do it is to show that $$ \int_{\mathbb{R}} \tilde{f_w}\tilde{\varphi}~dx = C\int_{\mathbb{R}}\xi\tilde{f}\tilde{\varphi}~d\xi $$ for every $\varphi\in\mathcal{S}(\mathbb{R})$. Note that the first integral makes sense for $\tilde{f_w}\in L^2$. You can justify this identity by Plancherel applied to the definition of weak derivative.

2. The norms being defined by a sum is not a problem, because these are $L^2$ norms and you're summing the square of the norms. So $$ \|f\|_2^2 + \|g\|_2^2 = \int_{\mathbb{R}} (|f|^2 + |g|^2)~dx. $$ Also, keep in mind that you are free to lose a multiplicative constant. You are not being asked to show that the norms are equal, merely equivalent up to constants.