Problem 1: 10pts
Compute the Fourier transform of the function $u(x)=\frac 1{1+x^4}\in L^1(\mathbb R)$, that is, for any $\xi\in\mathbb R$, compute the following integral
$$\hat u(\xi)=\int_{\mathbb R} \frac{e^{-i x\xi}}{1+x^4}\mathrm dx.$$
Construct a semicircle in the upper half-plane $\{\mathrm{Im}(z)\geq 0\}$ of radius $R>0$ and center at the origin, and denote it by $C_R$. Consider the closed contour $\Gamma_R=C_R\cup[-R,R]$. Then, when $\xi\leq 0$, we have in counterclockwise orientation
$$\hat u(\xi)=\lim_{R\to\infty}\int^{R}_{-R} \frac{e^{-i x\xi}}{1+x^4}\mathrm dx=\lim_{R\to\infty}\int_{\Gamma_R} \frac{e^{-i z\xi}}{1+z^4}\mathrm dz-\lim_{R\to\infty}\int_{C_R} \frac{e^{-i z\xi}}{1+z^4}\mathrm dz.$$The roots of the polynomial $1+z^4$ are given by
$$z_1=e^{i\pi/4},\quad z_2=e^{3i\pi/4},\quad z_3=e^{5i\pi/4},\quad z_4=e^{7i\pi/4}.$$As $R$ is sufficiently large, the only poles of the integrand inside $\Gamma_R$ are $z_1$ and $z_2$. By the residue theorem, we have
$$\int_{\Gamma_R} \frac{e^{-i z\xi}}{1+z^4}\mathrm dz=2\pi i\left[\mathrm{Res}\left(\frac{e^{-i z\xi}}{1+z^4},z_1\right)+\mathrm{Res}\left(\frac{e^{-i z\xi}}{1+z^4},z_2\right)\right],$$where
$$\mathrm{Res}\left(\frac{e^{-i z\xi}}{1+z^4},z_1\right)=\lim_{z\to z_1} (z-z_1)\frac{e^{-i z\xi}}{1+z^4}=\lim_{z\to z_1} \frac{e^{-i z\xi}}{4z^3}=\frac{e^{-i z_1\xi}}{4z_1^3},$$and the analogous result holds for $z_2$. Now it turns to estimate the integral over the semicircle $C_R$. We have
$$\left|\int_{C_R} \frac{e^{-i z\xi}}{1+z^4}\mathrm dz\right|\leq \sup_{z\in C_R}\left|\frac{e^{-i z\xi}}{1+z^4}\right|\cdot \mathrm{length}(C_R)\leq \frac{\pi R}{R^4-1}\to 0,\quad R\to\infty.$$Finally, we have
$$\hat u(\xi)=2\pi i\left[\frac{e^{-i z_1\xi}}{4z_1^3}+\frac{e^{-i z_2\xi}}{4z_2^3}\right]=\dfrac {\sqrt 2\pi }2e^{\frac {\sqrt 2}2\xi}\left[\cos\left(\frac {\sqrt 2}2\xi\right)-\sin\left(\frac {\sqrt 2}2\xi\right)\right],\quad \xi\leq 0.$$As $u(x)$ is an even function, we have $\hat u(\xi)=\hat u(-\xi)$, and thus
$$\hat u(\xi)=\dfrac {\sqrt 2\pi }2e^{-\frac {\sqrt 2}2|\xi|}\left[\cos\left(\frac {\sqrt 2}2|\xi|\right)+\sin\left(\frac {\sqrt 2}2|\xi|\right)\right],\quad \xi\in\mathbb R.$$Problem 2: 10pts
Find the solution to the initial value problem for $u=u(x)$,
$$x\dfrac {\mathrm d^2u}{\mathrm dx^2}-2(x+1)\dfrac {\mathrm du}{\mathrm dx}+(x+2)u=0,\quad u(1)=e,\ u'(1)=2e.$$(Hint: First check that $u=e^x$ is a solution to the equation.)
Firstly, we check that $u=e^x$ is a solution to the equation as follows
$$x\dfrac {\mathrm d^2e^x}{\mathrm dx^2}-2(x+1)\dfrac {\mathrm d e^x}{\mathrm dx}+(x+2)e^x=e^x(x-2x-2-x-2)=0,$$and $u=e^x$ is a particular solution to the equation. Then let $u(x)=v(x)e^x$, by variation of parameters, we have that $v(x)$ satisfies the following equation
$$xv''-2v'=0.$$Solving this equation, we get
$$v(x)=C_1x^3+C_2,$$where $C_1$ and $C_2$ are constants. Thus, the general solution to the equation is given by
$$u(x)=(C_1x^3+C_2)e^x.$$To satisfy the initial conditions, we have
$$u(x)=\left(\dfrac 13x^3+\dfrac 23\right)e^x.$$Problem 3: 10pts
Prove that the following limit exists in $\mathcal D'(\mathbb R)$ (in the sense of distribution on $\mathbb R$), and find the limit:
$$\lim_{n\to +\infty}n^2|x|\cos(nx).$$
To find the limit of $n^2|x|\cos(nx)$ in the sense of distributions, we take an abitrary test function $\phi\in \mathcal D(\mathbb R)$, that is we need to calculate the following limit
$$I=\lim_{n\to+\infty}\langle n^2|x|\cos(nx),\phi\rangle=\lim_{n\to+\infty}\int_{\mathbb R} n^2|x|\cos(nx)\phi(x)\mathrm dx.$$By the evenness of $n^2|x|\cos(nx)$, we denote $\psi(x)=\phi(x)+\phi(-x)$, and we have
$$I=\lim_{n\to+\infty}\int_0^{+\infty} n^2x\cos(nx)\psi(x)\mathrm dx,\quad \psi\in \mathcal D(\mathbb R).$$Integrating by parts, note that $\psi(x)$ has compact support, we have
$$I=-\lim_{n\to+\infty}n\int_0^{+\infty}(\psi(x)+x\psi'(x))\sin(nx)\mathrm dx.$$Analogously, we can integrate by parts again, and we have
$$I=\lim_{n\to+\infty}(\psi(x)+x\psi'(x))\cos(nx)\Big|_0^{+\infty}-\int_0^{+\infty}(2\psi'(x)+x\psi''(x))\cos(nx)\mathrm dx.$$As $2\psi'(x)+x\psi''(x)\in \mathcal D(\mathbb R)$, by the Riemann-Lebesgue lemma, such limit of the integral is zero. Finally,
$$I=\lim_{n\to+\infty}(\psi(x)+x\psi'(x))\cos(nx)\Big|_0^{+\infty}=-\psi(0)=-2\phi(0).$$That is, we have the limit in the sense of distributions
$$\lim_{n\to +\infty}n^2|x|\cos(nx)=-2\delta.$$Problem 4: 10pts (incorrect)
Let $X,Y$ be Banach spaces. Let $T:X\to Y$ be a bounded linear map. Suppose that $T$ maps every bounded closed set to a closed set. Prove that $T(X)$ is closed in $Y$.
Firstly we assume that $T$ is injective. To show $T(X)$ is closed in $Y$, we only need to show that for any Cauchy sequence $\{y_n\}\subset T(X)$, the limit of it, denoted by $y$, is also in $T(X)$. As $T$ is injective, there exists a unique sequence $\{x_n=T^{-1}(y_n)\}\subset X$. If $\{x_n\}$ is bounded, then $\{x_n\}'\neq \varnothing$, and we obtain a convergent subsequence $x_{n_k}\to x\in X$, so that $y_{n_k}=T(x_{n_k})\to T(x)=y\in T(X)$. If $\{x_n\}$ is unbounded, we claim that $\|x_n\|\to +\infty$. Otherwise, there exists a subsequence $\{x_{n_k}\}$ such that $\|x_{n_k}\|$ is bounded, by the given condition, $T(\{x_{n_k}\})$ is closed, and thus $y\in T(X)$ by the previous argument. Now $\|x_n\|\to +\infty$, as $T$ is bounded, we have
$$\|y_n\|=\|T(x_n)\|\leq \|T\|\cdot \|x_n\|.$$Consider the sequence
$$z_n=\frac{x_n}{\|x_n\|},\quad \|z_n\|=1.$$Then $\{z_n\}$ is bounded, and thus there exists a convergent subsequence $z_{n_k}\to z\in S_X$, where $S_X$ is the unit sphere in $X$. By the boundedness of $T$, we have
$$T(z)\leftarrow T(z_{n_k})=\frac{y_{n_k}}{\|x_{n_k}\|}\to 0.$$That is $T(z)=0$, which contradicts the injectivity of $T$.
Now we consider the general case where $T$ is not injective, then we can consider the quotient space $X/\ker(T)$, which is also a Banach space. And
$$\tilde T:X/\ker(T)\to Y,\quad \tilde T(x+\ker(T))=T(x)$$is a well-defined injective bounded linear map. It remains to show that $\tilde T$ maps every bounded closed set to a closed set. Let $A\subset X/\ker(T)$ be a bounded closed set, then by the continuity of the quotient map $\pi:X\to X/\ker(T)$, we have that
$$\pi^{-1}(A)=\{x\in X:x+\ker(T)\in A\}.$$Consider
$$F=\{x\in X:[x]\in A\}\cap \{x\in X:\|x\|\leq M+1\},$$where $M$ is the bound of $A$. Then $\pi(F)=A$ and $F$ is bounded and closed in $X$. By the given condition, $T(F)$ is closed in $Y$, and thus $\tilde T(A)=T(F)$ is closed in $Y$.
Finally, we solve all the cases.
Problem 5: 15pts (incomplete)
Let $\Omega\subseteq \mathbb C$ be open and connected, and $f_n:\Omega\to \mathbb C$ be holomorphic(i.e., analytic) for each $n\in\mathbb N$. Suppose $f_n$ converges to $f$ uniformly on each compact subset of $\Omega$, and $f$ is non-constant. Let $U\subseteq \Omega$ be open. For each compact subset $K\subseteq f(U)$, prove that $K\subseteq f_n(U)$ for all sufficiently large $n$.
As $f$ is holomorphic, for each compact subset $K\subseteq f(U)$, we have that $f^{-1}(K)$ is compact in $U$.
Problem 6: 15pts (more about AC)
Let $f:\mathbb R\to\mathbb R$ be locally integrable (i.e. Lebesgue integrable on any compact interval). Assume that $f$ is weakly differentiable, that is, its weak derivative $f'$ (in the sense of distribution) is also a locally integrable function. Prove that, except for a zero measure set, $f$ agrees with a function $g:\mathbb R\to\mathbb R$ that is absolutely continuous on any compact interval.
For any compact interval $[a,b]$, we define a function $g:[a,b]\to\mathbb R$ by
$$g(x)=\dfrac 1{b-a}\int_a^b f(t)\mathrm dt+\int_a^x f'(t)\mathrm dt.$$Then $g$ is absolutely continuous on $[a,b]$ by the fundamental theorem of calculus for Lebesgue integrals. Now we show that $f$ agrees with $g$ almost everywhere on $[a,b]$. It suffices to show that $h=f-g$ has a weak derivative $h'=0$ in the sense of distribution. And it’s easy to check
$$\langle h',\phi\rangle=-\langle h,\phi'\rangle=-\langle f-g,\phi'\rangle=-\langle f,\phi'\rangle+\langle g,\phi'\rangle=-\langle f',\phi\rangle+\langle f',\phi\rangle=0.$$Besides, we need to show that $g$ is independent of the choice of the compact interval $[a,b]$. As we have shown that $f$ agrees with $g$ almost everywhere on $[a,b]$, then for any other compact interval $[c,d]$, we have that $f$ agrees with $g$ almost everywhere on $[c,d]$. Thus, $g$ is independent of the choice of the compact interval.
Problem 7: 5pts+10pts
Given $f\in C^1([0,\infty))$, consider the linear differential equation for $u=u(x)$ on $[0,\infty)$,
$$\dfrac {\mathrm d^2u}{\mathrm dx^2}-(1+f(x))u=0.$$(1) If $\lim_{x\to +\infty}f(x)=0$, prove that any non-zero solution $u=u(x)$ has at most finite number of zeroes on $[0,\infty)$.
(2) If $\int^\infty_0|f(x)|\mathrm dx<+\infty$, prove that there is a unique solution $u=u(x)$ satisfying
$$\lim_{x\to +\infty}e^xu(x)=1.$$
(1) We claim that all zeroes of $u$ are isolated. Otherwise, there exists a sequence $\{x_n\}\subset [0,\infty)$ such that $u(x_n)=0$ and $x_n\to x_0\in[0,\infty)$. By the continuity of $u$, we have
$$u(x_0)=\lim_{n\to\infty}u(x_n)=0,\quad u'(x_0)=\lim_{n\to\infty}\dfrac {u(x_n)-u(x_0)}{x_n-x_0}=0.$$Then by the Picard-Lindelöf theorem, we have $u\equiv 0$, which contradicts the assumption that $u$ is non-zero. Thus, all zeroes of $u$ are isolated. As $\lim_{x\to +\infty}f(x)=0$, there exists $M>0$ such that $|f(x)|<1/2$ for all $x>M$. If there are two zeroes $x_1,x_2>M$ of $u$, then by isolation of zeroes, we assume that $u(x)>0$ for all $x\in(x_1,x_2)$, then
$$u''(x)=(1+f(x))u(x)>0,\quad \forall x\in(x_1,x_2),$$which implies that $u'$ is strictly increasing on $(x_1,x_2)$. Thus
$$0=u'(x_1)