I think that I am begginer of PDE. I am reading the Evan's PDE, p.282, Theorem 7 and some question arises :
Q.1. First underlined statement : Where did $C$ come from? In previous sentence Evans wrote that "In particular choosing $\alpha= ( k, 0 , \dots , 0 ), (0,k, \dots,0), \dots, (0, \dots , k )$, ". Why such choice allows us to deduce the first underlined inequality?
Q.2. Second underlined statement : Likewise, where did $C$ come from? And in procedure of deriving the $(18)$ from the $(20)$, $||u||_{H^{k}(\mathbb{R}^n)} \le ||(iy)^{\alpha}\hat{u}||_{L^2(\mathbb{R}^n)}$? If so, why?
Q.3. Final underlined equality : Why the equality is true? How can we derive the equality? My one trial, by using Theorem 4.3.2. ( p. 181 ) in the Evans's book :
Theorem 2. ( Properties of Fourier transform ). Assume $u,v \in L^2(\mathbb{R}^n)$. Then
- (i) $\int_{\mathbb{R}^n} u \bar{v} dx =\int_{\mathbb{R}^n}\hat{u}\bar{\hat{v}} dy$,
- (ii) $\hat{D^{\alpha}u}=(iy)^{\alpha}\hat{u}$ for each multiindex $\alpha$ such that $D^{\alpha}u \in L^{2}(\mathbb{R}^{n})$,
- (iii) $( u *v)^{\wedge} =(2\pi)^{n/2}\hat{u}\hat{v}$,
- (iv) $u = (\hat{u}\check) $.
, is as follows ( We omit $\mathbb{R}^{n}$, which is the domain of integration ) :
$$(-1)^{|\alpha|} \int \phi \bar{u_{\alpha}} dx = (-1)^{|{\alpha}|} \int \hat{\phi} \bar{\hat{u_{\alpha}}} dy := (-1)^{|{\alpha}|} \int \hat{\phi} \overline{ (((iy)^{\alpha} \hat{u})\check))^{\wedge} } dy
\stackrel{?}{=} (-1)^{|\alpha|} \int \hat{\phi} \overline{ (((iy)^{\alpha})\check) ((\hat{u})\check) )^{\wedge} }dy \stackrel{?}{=} (-1)^{|\alpha|} \int \hat{\phi} \overline{ ((iy)^{\alpha})\check)^{\wedge} ( ( \hat{u})\check)^{\wedge}}
\stackrel{?}{=} (-1)^{|\alpha|}\int \hat{\phi} \overline{(iy)^{\alpha} \hat{u}}dy = (-1)^{|\alpha|}\int \hat{\phi} \overline{(iy)^{\alpha}}\bar{\hat{u}} dy \stackrel{?}{=} \int (iy)^{\alpha} \hat{\phi}\bar{\hat{u}}dy $$
Each steps marked by the question symbols are true? Third and Fourth equality : The Fourier and its inverse transform is distributive under the multiplication ? Fifth equality : $ ( (iy)^{\alpha} \check)^{\wedge}=(iy)^{\alpha} $? By the above theorem $2$, for $u\in L^{2}(\mathbb{R}^n)$, $u = (\hat{u}\check) $ . Perhaps, $ ( u \check)^{\wedge} = u$ is also true? If it is not true in general, how about for $u=(iy)^{\alpha}$? Final equality : To show the final equality, perhaps, $(-1)^{|\alpha|} \overline{(iy)^{\alpha}} = (iy)^{\alpha}$?
I don't understand these three questions at all until now. Can anyone explain those more friendly?
Best Answer
I'll preface this answer by saying that these types of question can be hard to answer in a satisfactory manner since it's not always easy to put yourselves in the author's shoes, so please excuse me if I don't understand or do exactly what Evans probably intended, especially for your first question.
I'm not exactly sure myself why Evans chooses these indexes in particular and not all of those such that $|\alpha| = k$, but it could maybe be a typo in the equation below, with $|y|^{2k}$ where it could have been $|y_j|^{2k}$. Indeed, if it had been $|y_j|^{2k}$, then it would make sense to choose $\alpha = (0, \cdots, 0, k, 0, \cdots, 0)$ with $k$ in the $i$-th position.
Either way, the overall idea is to use a few equalities and inequalities related to multi-indices.
For example, we have: $$\left(\sum_{j = 1}^n x_j\right)^k = \sum_{|\alpha| = k} \binom{k}{\alpha} x^\alpha$$ where $$\binom{k}{\alpha} := \frac{k!}{\prod_{j = 1}^n \alpha_j!}$$ This is a generalisation of the binomial theorem called the multinomial theorem.
If you apply it to $x_j = |y_j|^{2}$, you get: $$|y|^{2k} = \left(\sum_{j = 1}^n |y_j|^2\right)^k = \sum_{|\alpha| = k} \binom{k}{\alpha} \prod_{j = 1}^n |y_j|^{2\alpha_j} = \sum_{|\alpha| = k} \binom{k}{\alpha} |y^\alpha|^2$$ so that: $$\begin{split}\int_{\mathbb{R}^n} |y|^{2k}|\hat{u}|^2\,\mathrm{d}y &= \int_{\mathbb{R}^n} \left(\sum_{|\alpha| = k} \binom{k}{\alpha} |y^\alpha|^2\right) |\hat{u}|^2\,\mathrm{d}y\\ &\leq \left(\sup_{|\alpha| = k} \binom{k}{\alpha}\right) \int_{\mathbb{R}^n} \left(\sum_{|\alpha| = k}|y^\alpha \hat{u}|^2\right)\,\mathrm{d}y\\ &\leq \left(\sup_{|\alpha| = k} \binom{k}{\alpha}\right) \int_{\mathbb{R}^n} \left(\sum_{|\alpha| = k}|D^\alpha u|^2\right)\,\mathrm{d}y =: C_k \int_{\mathbb{R}^n} |D^k u|^2\,\mathrm{d}y\end{split}$$
Here we have the same kind of deal. This is a trick that's pretty basic but quite helpful: instead of trying to prove the inequality $$|y|^{2|\alpha|} \leq C \left(1 + |y|^k\right)^2$$ via some direct upper-bounding, you instead study the function $f : t \in \mathbb{R} \mapsto \displaystyle\frac{t^{|\alpha|}}{1 + t^k}$ and try to show it is bounded by relying on its continuity, and then apply that to $t = |y|$.
It's pretty clear that the limit at $+\infty$ of $f$ is $0$ if $|\alpha| < k$ and $1$ if $|\alpha| = k$, $f$ is non-negative, and $f$ is continuous on $[0, +\infty)$, therefore (left as a quick exercise) $f$ is bounded on $[0,+\infty)$, and we have the desired inequality.
You are on the right track but also you are overcomplicating the problem a tiny bit. Indeed the only thing you need to prove is: $$\overline{(iy)^\alpha} = (-1)^{|\alpha|}(iy)^\alpha$$ Once this is proven true, you'll be able to proceed like this: $$\int (iy)^{\alpha} \hat{\phi}\bar{\hat{u}}\,\mathrm{d}y = (-1)^{|\alpha|}\int \hat{\phi} \overline{(iy)^{\alpha}\hat{u}}\,\mathrm{d}y = (-1)^{|\alpha|}\int \hat{\phi} \overline{\widehat{u_\alpha}}\,\mathrm{d}y = (-1)^{|\alpha|}\int \phi \overline{u_\alpha}\,\mathrm{d}y$$ However, because all the components of $y$ are real, we have: $$\begin{split}\overline{(iy)^\alpha} &= \overline{\prod_{j = 1}^n (iy_j)^{\alpha_j}}\\ &= \prod_{j = 1}^n \left(\overline{iy_j}\right)^{\alpha_j}\\ &= \prod_{j = 1}^n \left(-iy_j\right)^{\alpha_j}\\ &= \left(\prod_{j = 1}^n \left(-1\right)^{\alpha_j}\right) \left(\prod_{j = 1}^n \left(iy_j\right)^{\alpha_j}\right)\\ &= (-1)^{|\alpha|} (iy)^\alpha\end{split}$$ and we have shown what we wanted.