Second assignment with solutions
was due September 18 in class
18.155 Fall 2001

Richard Melrose

Department of Mathematics, Massachusetts Institute of Technology

rbm@math.mit.edu

In the main these questions form theorems in Hörmander's book [1], so the proofs are available there. I suggest that you try to work them out on your own and in any case I expect written proofs, even if you need to get the idea from the book. Of course, at the very least, you will have to translate the notation.

Problem 1   [Hörmander, Theorem 3.1.4] Let $I\subset\mathbb{R}$ be an open, non-empty interval.

1. [i)] Show (you may use results from class) that there exists $\psi\in\mathcal{C}_c(I)$ with $\int_{\mathbb{R}}\psi(x)ds=1.$
   Proof. [Solution] We showed in class that there exists $\phi\in\mathcal{C}_c(\mathbb{R})$ which is strictly positive on $(0,1)$ and vanishes on $\mathbb{R}\setminus(0,1).$ If $I$ is a non-empty open interval it contains an interval $[\bar x-\epsilon,\bar x+\epsilon]$ for some $\bar x\in\mathbb{R}$ and $\epsilon>0.$ Consider $\psi(x)=C\phi((x-\bar x)/\epsilon)$ where $C>0$ is to be chosen. The properties of $\phi$ imply that $\psi\in\mathcal{C}_c(I)$ and that $\int_{\mathbb{R}}\psi=1/C,$ $C>0.$ This value of $C$ make the integral of $\psi$ equal $1$ as desired. $\qedsymbol$

2. [ii)] Show that any $\phi \in\mathcal{C}_c(I)$ may be written in the form

   $$\phi=\tilde\phi+c\psi,\ c\in\mathbb{C},\ \tilde\phi \in\mathcal{C}_c(I) \int_{\mathbb{R}}\tilde\phi =0.$$

   
   
   Proof. [Solution] Set $c=\int_{I}\phi$ and observe that $\tilde\phi=\phi-c\psi$ has vanishing integral. $\qedsymbol$

3. [iii)] Show that if $\tilde\phi\in\mathcal{C}_c(I)$ and $\int_{\mathbb{R}}\tilde\phi =0$ then there exists $\mu \in\mathcal{C}_c(I)$ such that $\frac{d\mu }{dx}=\tilde\phi$ in $I.$
   Proof. [Solution] If we consider $\mu(x)=\int_{-\infty}^x\tilde\phi(s)ds$ then $\mu$ is infinitely differentiable since $d\mu(x)/dx=\tilde\phi(x)$ by the fundamental theorem of calculus. Certainly $\mu(x)=0$ if $x<\inf I',$ where $I'$ is a compact interval outside which $\tilde\phi$ vanishes. The assumption that $\int_{I}\tilde\phi(x)dx=\int_{I'}\tilde\phi(x)dx=0$ means that $\mu(x)=0$ for $x>\sup I'$ so $\mu\in\mathcal{C}_c(I).$ $\qedsymbol$

4. [iv)] Suppose $u\in\mathcal{C}^{-\infty}(I)$ satisfies $\frac{du}{dx}=0,$ i.e.

   $$u(-\frac{d\phi}{dx})=0 \forall \phi \in\mathcal{C}_c(I),$$ (1)

   
   
   show that $u=c$ for some constant $c.$
   Proof. [Solution] Using ii) and iii) we may write each element $\phi \in\mathcal{C}_c(I)$ in the form $\phi=c\psi+d\mu(x)/dx$ where $\psi\in\mathcal{C}_c(I)$ is fixed and $\mu \in\mathcal{C}_c(I)$ depends on $\phi.$ Then the definition of the vanishing of the derivative in 1 shows that

   $$u(\phi)=u(c\psi+d\mu/dx)=cu(\psi)-u(-d\mu/dx)=cu(\psi)=\int_IC\phi(x)dx$$

   
   

   where we have used the value of $c$ and set $C=u(\psi).$ The last integral is the definition of the (constant) function $C$ as a distribution on $I.$ Thus $u=C$ is an equality between distributions. $\qedsymbol$

5. [v)] Suppose that $u\in\mathcal{C}^{-\infty}(I)$ satisfies $\frac{du}{dx}=c,$ for some constant $c,$ show that $u=cx+d$ for some $d\in\mathbb{C}.$
   Proof. [Solution] Consider the continuous function $cx.$ This defines a distribution, which we just denote the same way. The definition of the derivative means that $d(cx)/dx=c$ (as you would expect). We proved this in class, it is the integration by parts formula

   $$\frac{d(cx)}{dx}(\phi)=cx(-\frac{d\phi}{dx})=-\int_{\mathbb{R}}cx\frac{d\phi}{dx}dx =c(\phi).$$

   
   

   Now, if we consider the difference $v=u-cx$ it is a distribution on $I$ which satisfies $dv/dx=0,$ so by iv) it is equal to a constant, $d.$ That is $u-cx=d,$ or $u=cx+d.$ $\qedsymbol$

Problem 2   [Hörmander Theorem 3.1.16]

1. [i)] Use Taylor's formula to show that there is a fixed $\psi \in\mathcal{C}_c(\mathbb{R}^n)$ such that any $\phi \in\mathcal{C}_c(\mathbb{R}^n)$ can be written in the form

   $$\phi =c\psi +\sum\limits_{j=1}^nx_j\psi _j$$ (2)

   
   
   where $c\in\mathbb{C}$ and the $\psi _j\in\mathcal{C}_c(\mathbb{R}^n)$ depend on $\phi.$
   Proof. [Solution] By Taylor's formula, or theorem, any infinitely differentiable function which vanishes at 0 can be written in the form

   $$\phi=\sum\limits_{j=1}^nx_ju_j$$ (3)

   
   

   where the $u_j$ are also infinitely differeniable. Choosing $\psi\in\mathcal{C}_c(\mathbb{R}^n),$ as in class, with $\psi(0)=1$ we can apply this to $\phi-c\psi$ where $c=\phi(0).$ The problem is that the $u_j$ need not have compact support. However, we can choose another function (depending on $\phi),$ $\psi'\in\mathcal{C}_c(\mathbb{R}^n)$ which is equal to $1$ on the support of $\phi$ and the support of $\psi.$ Thus $\phi\psi'=\phi$ and $\psi\psi'=\psi.$ Then 3 gives the desired 2 where $\psi_j-u_j\psi'\in\mathcal{C}_c(\mathbb{R}^n).$ $\qedsymbol$

2. [ii)]Recall that $\delta _0$ is the distribution defined by

   $$\delta _0(\phi)=\phi(0)\ \forall\ \phi \in\mathcal{C}_c(\mathbb{R}^n);$$

   
   
   explain why $\delta _0\in\mathcal{C}^{-\infty}(\mathbb{R}^n).$
   Proof. [Solution] Certainly $\delta _0$ so defined is a linear functional and it is continuous since

   $$\vert\delta _0(\phi)\vert\le \sup_{\mathbb{R}^n}\vert\phi\vert=\Vert\phi\Vert _0.$$

   
   

   $\qedsymbol$

3. [iii)] Show that if $u\in\mathcal{C}^{-\infty}(\mathbb{R}^n)$ and $u(x_j\phi )=0$ for all $\phi \in\mathcal{C}_c(\mathbb{R}^n)$ and $j=1,\dots,n$ then $u=c\delta _0$ for some $c\in\mathbb{C}.$
   Proof. [Solution] We can use 2 to evaluate

   $$u(\phi)=u(c\psi)+\sum\limits_{j=1}^nu(x_j\psi_j)=C\phi(0)$$

   
   

   where the given identity means that the sum is zero and we write $C=u(\psi)$ for a fixed constant and note that $c=\phi(0).$ This just means $u=C\delta_0.$ $\qedsymbol$

4. [iv)] Define the `Heaviside function'

   $$H(\phi )=\int_0^\infty\phi (x)dx\ \forall\ \phi \in\mathcal{C}_c(\mathbb{R});$$

   
   
   show that $H\in\mathcal{C}^{-\infty}(\mathbb{R}).$
   Proof. [Solution] The linearity of the integral shows that $H$ is a linear functional on $\mathcal{C}_c(\mathbb{R}).$ The basic integral estimate

   $$\vert H(\phi)\vert\le L\sup\vert\phi\vert \phi(x)=0 \vert x\vert>L$$

   
   

   shows that it is continuous. $\qedsymbol$

5. [v)] Compute $\frac{d}{dx}H\in\mathcal{C}^{-\infty}(\mathbb{R}).$
   Proof. [Solution] By definition

   $$\frac{dH}{dx}(\phi)=-H(\frac{d\phi}{dx})=-\int_0^\infty\frac{d\phi}{dx}dx=\phi(0)$$

   
   

   where we have used the fundamental theorem of calculus. Thus

   $$\frac{dH}{dx}=\delta_0.$$ (4)

   
   

   $\qedsymbol$

Problem 3   Using Problems 1 and 2, find all $u\in\mathcal{C}^{-\infty}(\mathbb{R})$ satisfying the differential equation

$$x\frac{du}{dx}=0 \mathbb{R}.$$

Proof. [Solution] By definition $v=\frac{du}{dx}$ is a distribution which satisfies $xv=0.$ Problem 2, part iii), in one dimension shows that $v=c\delta_0.$ Thus $\frac{du}{dx}=c\delta_0.$ By Problem 2, part v), $dH/dx=\delta.$ Thus $\tilde u=u-cH$ satisfies $d\tilde u/dx=0.$ By Problem 1, part iv), we conclude that $v=d$ is constant. Thus the only distributional solutions of $x\frac{du}{dx}=0$ are

$$u=cH+d,\ c d$$

This means that $u$ is constant in $x<0$ and in $x>0,$ with possibly different values, but it means more than this as regards its `behaviour at 0'. $\qedsymbol$