Let $P\in L^2(\Omega)$ (where $\Omega=[0,2\pi]^n$), be a function such that : $\forall x=(x_1,...,x_n)\in \Omega,\;P(x)=\sum_{j_1=-\infty}^{+\infty}...\sum_{j_n=-\infty}^{+\infty}a_{j}e^{ij.x}$ with $j.x=j_1x_1+...+j_nx_n$.

How can I figure out $$\left \| P \right \|_{H^{-1}(\Omega)}\; \text{and}\;\; \left \| \nabla P \right \|_{(H^{-1}(\Omega))^n}\;,$$ I will appreciate if you could help me . Thanks.

PS: I want to prove that:

$$\left \| P \right \|_{L^{2}(\Omega)}\; \leq c\big(\left \| P \right \|_{H^{-1}(\Omega)}+\; \left \| \nabla P \right \|_{(H^{-1}(\Omega))^n}\;\big).$$

Directly using the calculations, and without using any particular inequality (in particular the inequality of $Ne\check{c}as$.)

Here what I did: This is not an answer but it's the idea that i have, if some one can complete it:

Let's try to characterize $\left \| P \right \|_{H^{-1}(Q)}$ and $\left \| \nabla P \right \|_{(H^{-1}(Q))^d}$. We have : $$\left \| P \right \|_{H^{-1}(Q)}\geqslant \frac{\left \langle P,\phi \right \rangle}{ \left \| \phi \right \|_{H^1(Q)}}\;,\:\: \forall \phi \in H_0^1(Q)$$ In particular for each $N\in \mathbb N$, Let $$\phi_N(x)=\sum_{j\in[-N,N]^d} (1+\left | j \right |^2)^{-1}\;\overline{a_j}\;e^{i\,j.x} \;,\;\forall x\in Q$$ We have: \begin{align*} \left \| \phi_N \right \|_{H^1(Q)}&=(\sum_{j\in[-N,N]^d }\; (1+\left | j \right |^2)(1+\left | j \right |^2)^{-2}\;\left | a_j \right |^2)^\frac{1}{2} \\ &= (\sum_{j\in[-N,N]^d}\; (1+\left | j \right |^2)^{-1}\;\left | a_j \right |^2)^\frac{1}{2} \end{align*} So, \begin{align*} \left \| P \right \|_{H^{-1}(Q)} &\geqslant \frac{\left \langle P,\phi_N \right \rangle}{ \left \| \phi_N \right \|_{H^1}}\\ &=\frac{\sum_{j\in[-N,N]^d}\; (1+\left | j \right |^2)^{-1}\;\left | a_j \right |^2}{(\sum_{j\in[-N,N]^d}\; (1+\left | j \right |^2)^{-1}\;\left | a_j \right |^2)^\frac{1}{2}}\\ &=(\sum_{j\in[-N,N]^d}\; (1+\left | j \right |^2)^{-1}\;\left | a_j \right |^2)^\frac{1}{2}\;,\;\forall N\in\mathbb N \end{align*} So, $\left \| P \right \|_{H^{-1}(Q)} \geqslant (\sum_{j\in\mathbb Z^d}\; (1+\left | j \right |^2)^{-1}\;\left | a_j \right |^2)^\frac{1}{2}$

I want to use the same idea (Find $\phi _N$) to prove that: $\left \| \nabla P \right \|_{(H^{-1}(Q))^d} \geqslant (\sum_{j\in\mathbb Z^d}\; \left | j \right |^2(1+\left | j \right |^2)^{-1}\;\left | a_j \right |^2)^\frac{1}{2}$ Can you help me?


Extending your $P$ by $0$ outside of $\Omega$. Now from Plancherel's equality and Hölder's inequality we should obtain that for a function $Q\in H_0^1$ \begin{equation} |P\cdot Q|=|\hat{P}\cdot\hat{Q}|\leq \|\frac{1}{1+x^2}\hat{P}\|_{L^2(\mathbb{R}^n)}\|(1+x^2)\hat{Q}\|_{L^2(\mathbb{R}^n)}, \end{equation} where the symbol $\hat{}$ means the Fourier transformation (and I just omit the corresponding constant here). This means that $\|P\|_{H^{-1}}\leq \|\frac{1}{1+x^2}\hat{P}\|_{L^2(\mathbb{R}^n)}$. But in fact, we can show that $\|P\|_{H^{-1}}$ is equivalent to $\|\frac{1}{1+x^2}\hat{P}\|_{L^2(\mathbb{R}^n)}$: let $P_n\in C_0^\infty$ be a sequence which converges to $P$ in $L^2$. Then define $Q_n$ be the fourier inverse of $\frac{\hat{P}_n}{(1+x^2)^2}$. From Paley-Wiener theorem we know that $Q_n$ also has compact support in $\Omega$, therefore $Q_n\in H^1_0(\Omega)$ with $\|Q_n\|_{H^1}=\|(1+x^2)\hat{Q}_n\|_{L^2}=\|\frac{\hat{P}_n}{1+x^2}\|_{L^2}$. Define $K_n:=Q_n/\|(1+x^2)\hat{Q}_n\|_{L^2}$, inserting this to $|P\cdot Q|$ with $Q=K_n$ we should obtain that $|P\cdot K_n|\leq \|P\|_{H^{-1}}$ and $|P\cdot K_n|$ converges to $\|\frac{\hat{P}_n}{1+x^2}\|_{L^2}$. This shows our claim.

Analogously we obtain that \begin{equation} |P\cdot \mathrm{div}Q|=|\hat{P}\cdot x^2\hat{Q}|\leq \|\frac{x^2}{1+x^2}\hat{P}\|_{L^2(\mathbb{R}^n)}\|(1+x^2)\hat{Q}\|_{L^2(\mathbb{R}^n)}, \end{equation} therefore $\|\nabla P\|_{H^{-1}}$ is equivalent to $\|\frac{x^2}{1+x^2}\hat{P}\|_{L^2(\mathbb{R}^n)}$. Now let $Q\in L^2$. We obtain that \begin{align} &|P\cdot Q|\\ =&|\hat{P}\cdot \hat{Q}|\\ =&|(\frac{\hat{P}}{1+x^2}+\frac{x^2\hat{P}}{1+x^2})\cdot \hat{Q}|\\ \leq& (\|\frac{1}{1+x^2}\hat{P}\|_{L^2(\mathbb{R}^n)}+\|\frac{x^2}{1+x^2}\hat{P}\|_{L^2(\mathbb{R}^n)})\|\hat{Q}\|_{L^2(\mathbb{R}^n)}\\ \leq &C(\|P\|_{H^{-1}}+\|\nabla P\|_{H^{-1}})\|\hat{Q}\|_{L^2(\mathbb{R}^n)}. \end{align} This completes the proof.

If you really want a so called "direct calculation", I would like to give the following suggest. Notice it is not a proof, and I don't know if it is correct, but at least it should look similar to my original proof and one may find a similar way to prove it: if $P$ is in $L^2$, then $\sum |a_{j_n}|^2<\infty$. Now a function $\sum b_{j_n}e^{ij_nx_n}$ is in $H^1$ if and only if $\sum(1+j^2_n)|b_{j_n}|^2<\infty$. If you now consider $P\cdot Q$, you will find it is equal to $\sum a_{j_n}b_{j_n}=\sum \frac{1}{1+j_n^2}a_{j_n}(1+j_n^2)b_{j_n}$. From here you should see that the proof should be in fact similar to my original proof.

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    $\begingroup$ Since the domain is a cube with sides of length $2\pi$, I would think that it is probably better to use the periodic setting of Fourier-series. Although the actual calculation will probably look similar. $\endgroup$ – mlk May 22 '17 at 14:58
  • $\begingroup$ @mlk: The writer of $P$ comes from this fact ^^. $\endgroup$ – Motaka May 22 '17 at 15:02
  • $\begingroup$ @Cuteboy: Thank you so much for your reply, but i need a direct calculation, I did not see where you used the fact that $P$ has a particular form .. $\endgroup$ – Motaka May 22 '17 at 15:12
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    $\begingroup$ I dont really understand what you mean a direct calculation o.o $\endgroup$ – ehochix May 22 '17 at 17:26
  • $\begingroup$ @Cuteboy: I mean by "direct calculation", a demonstration using the particular writing of p, like what you wrote in the end of your answer $\left \| P \right \|_2=\sum |a_{j_n}|^2<\infty$, and $\left \| P \right \|_{H^{-1}}=..$. $\endgroup$ – Motaka May 24 '17 at 10:15

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