PhD Qualifying Exam for Real Analysis, Spring 2013-14 by Dr Li

I just solved all of them except for 4. As a usual practice answer of qualifying exam will not be unveiled (even to PG students in UST), let me record my solution below for future use.

The following is the set of problems a few days ago:

Problem 1. Let $F$ be a Lebesgue nonmeasurable subset of $[0,1]$. Prove that there is $c\in (0,1)$ such that whenever $E\subseteq[0,1]$ is Lebesgue measurable and $m(E)\ge c$, then $F\cap E$ is Lebesgue nonmeasurable.

Problem 2. Let $r_1,r_2,r_3,\dots$ be a sequence containing each rational number in $[0,1]$ exactly once. Let $f:[0,1]\setminus \{r_1,r_2,r_3\dots\}\to \R$ be defined by $\dis f(x)=\sum_{k=1}^\infty \frac{1}{k^2|x-r_k|^{1/2}}$. Prove that $f(x)<\infty$ a.e. on $[0,1]\setminus \{r_1,r_2,r_3\dots\}$.

Problem 3. Let $S$ be a vector subspace of $L^2[0,1]$ with $\|f\|=(\int_{[0,1]}|f|^2\,dm )^{1/2}$ and suppose that there is a constant $K$ such that for every $f\in S$ and every $x\in [0,1]$, $|f(x)|\leq K\|f\|$. Show that the dimension of $S$ is finite.

My Remark. The converse is also true, and this can be generalized to every $L^p(K)$, where $p\in (1,\infty)$ and $K$ is compact.

Problem 4. For $f:[0,1]\to (0,\infty)$, if $f,\ln f,f\ln f$ are Lebesgue integrable on $[0,1]$, then find which one of the two numbers $\dis \int_{[0,1]}f\ln f\,dm$ and $\dis \bigg( \int_{[0,1]}f\,dm \bigg) \bigg(\int_{[0,1]}\ln f\,dm \bigg)$ is larger. Show all details.

Problem 5. Let $a_0,a_1,a_2,\dots$ be a sequence in $\R$ such that for all $b_0,b_1,b_2,\dots\in \R$ and for every $k=0,1,2,\dots$, we have $\dis \abs{\sum_{n=0}^k a_n b_n}^2\leq \int_0^1\abs{\sum_{n=0}^k b_nx^n}^2\,dx$. Prove that there exists a unique $f\in L^2[0,1]$ such that for all $n=0,1,2,\dots$, we have $a_n=\dis \int_{[0,1]}x^nf(x)\,dm$.

Problem 6. For $n=1,2,3,\dots$, let $f_n:[0,1]\to \R$ be absolutely continuous and $f_n(0)=0$. If $f_n'$ is a Cauchy sequence in $L^1[0,1]$ with Lebesgue measure, then prove that $f_n$ converges uniformly on $[0,1]$ to an absolutely continuous function.

 
Problem 7. Let $S_1,S_2,S_3,\dots$ be a sequence of closed sets in $\R^2$. It is known that for every $x\in \R$, there exists a positive integer $k$ such that $(x,0)\in S_k$. Prove that there exists a positive integer $k$ such that the intersection of $S_k$ with the real axis contains a nonempty open interval of the real axis. 

Problem 8. Let $f:\R\to (0,\infty)$ be Lebesgue measurable. Prove that for every $\varepsilon>0$, there exists $g:\R\to (0,\infty)$ Lebesgue measurable such that $\|f-g\|_{\infty}<\varepsilon$ and for all $r>0$, $m\{x:g(x)=r\}=0$.

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Solution to 1. Suppose on the contrary that for every $c\in (0,1)$, there is a measurable $E\subseteq [0,1]$ with $m(E)\ge c$ such that $F\cap E$ is Lebesgue measurable.

In particular, let's, for every $n$, choose $c=1-1/n$, then there is a measurable $E_n\subseteq [0,1]$ and $m(E_n)\ge 1-1/n$ such that $F\cap E_n$ is Lebesgue measurable. Let $E=\cupp_{n=1}^\infty E_n$, then $m(E)=1$, and both $E,F\cap E$ are Lebesgue measurable.

Note that $F=(F\cap E)\sqcup (F\setminus E)$, but $$m^*(F\setminus E)\leq m^*([0,1]\setminus E)=m([0,1])-m(E)=0,$$ therefore $F\setminus E$ has measure zero, and hence measurable. This implies $F$ is measurable, and contradiction.$\qed$

Solution to 2. It is enough to show $\dis \int_0^1 f(x)\,dx<\infty$, this is a direct consequence of monotone convergence theorem and a direction integration of $\dis \int_0^1\frac{1}{|x-r_k|^{1/2}}\,dx$.$\qed$

Solution to 3. My solution to this problem is long. (Please kindly share your view if a simpler solution is possible)

We will show that $\ol S$ is finite dimensional. For this, let's "extend" the inequality $|f(x)|\leq K\|f\|$ (for every $x\in [0,1]$) to $f\in \ol S$.

Suppose that $f\in \ol S$, then there is $\{g_n\}$ in $S$ such that $\|g_n-f\|\to 0$. Therefore $g_n(x)\to f(x)$ for a.e. $x\in [0,1]$, and thus $$|f(x)|\leftarrow |g_n(x)|\leq K\|g_n\|\to K\|f\|\quad \text{a.e..}$$ Although this inequality just holds a.e., this will be enough since later our key step will be to do integration on both sides to this inequality.

Now we show that the closed unit ball $B_{\ol S}$ in $\ol S$ is compact. Let $f_1,f_2,f_3,\dots \in \ol S$ be such that $\|f_n\|\leq 1$. Note that from hypothesis $|f(x)|\leq K$ a.e..

By $w$-sequential compactness of $L^2[0,1]$, there is a subsequence $\{f_{n_k}\}$ such that $f_{n_k}\rightharpoonup f$, and therefore $f_{n_k}(x)\to f(x)$ for every $x\in [0,1]$ since pointwise evaluation is a bounded linear functional on $L^2[0,1]$ by hypothesis.

Now by Egoroff Theorem, for every $\epsilon>0$ there is compact $E\subseteq [0,1]$ such that  $f_{n_k} \rightrightarrows f$ on $E$ and $m([0,1]\setminus E)<\epsilon$. Then for every $j,k$ and for  a.e. $x\in [0,1]$, $$\begin{align*} |f_{n_j}(x)-f_{n_k}(x)|&\leq K\|f_{n_j}-f_{n_k}\|\\ &=K\sqrt{\int_E |f_{n_j}-f_{n_k}|^2\,dm +\int_{[0,1]\setminus E}|f_{n_j}-f_{n_k}|^2\,dm } \\&\leq K\sqrt{\|f_{n_j}-f_{n_k}\|_E^2 +4K^2\epsilon}. \end{align*}$$
For the $\epsilon>0$ above we can find an $N$ such that $$j,k>N\implies |f_{n_j}(x)-f_{n_k}(x)|\leq K\sqrt{\epsilon+4K^2\epsilon}=K\sqrt{1+4K^2}\sqrt{\epsilon}\quad \text{a.e..}$$ This implies $$j,k>N\implies \|f_{n_j}-f_{n_k}\|\leq C\sqrt{\epsilon}$$ for some absolute constant $C$, so $\{f_{n_k}\}$ is Cauchy in $L^2[0,1]$, so it converges to a $g\in L^2[0,1]$, and hence necessarily $g=f$ a.e.. We conclude that
$$\|f_{n_k}-f\| \to 0.$$
Therefore every $\{f_n\}$ in $B_{\ol S}$ has a convergent subsequence in $\ol S$, $B_{\ol S}$ is compact, hence $\ol S$ is finite dimensional.$\qed$

Solution to 4. No Idea.

Solution to 5. Let $\varphi_n(x)=x^2\in C[0,1]\subseteq L^2[0,1]$. The problem asks us to find $f\in L^2[0,1]$ such that $f(\varphi_n) = a_n$. So, let's define $$T:\spann\{\varphi_n:n\ge 1\}\to \R$$ by defining $T(\varphi_n) = a_n$ and extending it linearly. Now we show that $T$ is a bounded linear functional. Indeed, by hypothesis, for every $b_1,b_2\dots\in \R$ we have $$\abs{\sum_{n=0}^k T(\varphi_n) b_n}^2\leq \int_0^1\abs{\sum_{n=0}^k b_nx^n}^2\,dx$$ which can be rewritten as $$\abs{T\brac{\sum_{n=0}^k b_n\varphi_n}}^2\leq \int_0^1 \abs{\sum_{n=0}^k b_n\varphi_n}^2\,dx=\left \|\sum_{n=0}^k b_n\varphi_n\right\|^2,$$ as desired.

By Hahn-Banach Theorem, we can extend $T$ to $\widehat T:L^2[0,1]\to \R$, then by Riesz Representation Theorem for Hilbert spaces, there is a unique $f\in L^2[0,1]$ such that $$\widehat T(g) =\int_0^1 g f\,dm$$ for every $g\in L^2[0,1]$. Set $g=\varphi_n$, then we have $$a_n=T(\varphi_n)=\widehat T(\varphi_n)=\int_0^1 \varphi_n f\,dm.$$
It remains to show such $f$ is unique. Suppose that $\int_0^1 x^n(f-g)\,dm=0$ for every $n\ge 0$, then by Weierstrass Approximation, $\int_0^1 u(f-g)\,dm=0$ for every $u\in C[0,1]$. Finally, this implies $f=g$ a.e. since $m$ is a regular measure.$\qed$

Solution to 6. We just need to recall that $f_n$'s are differentiable a.e. and has the formula $f_n(x)=\int_0^xf_n'(t)\,dt$ due to absolute continuity. The the rest follows from completeness argument several times.$\qed$

Solution to 7. The hypothesis says that $\R\times \{0\}=\cupp_{k=1}^\infty (S_k\cap (\R\times \{0\}))$. Since $\R\times \{0\}$ is a closed subspace of $\R$, it must be complete. By Baire Category Theorem, there is a $k$ such that $B_{\R\times \{0\}} (a,\delta) = B(a,\delta)\cap (\R\times \{0\})\subseteq S_k \cap (\R\times \{0\})$ for some $a\in \R,\delta>0$.$\qed$

Solution to 8. Let $\varepsilon>0$, then define $\delta_n=n\frac{\varepsilon}{2}$, the intervals $(\delta_n,\delta_{n+1}]$'s form a partition of $(0,\infty)$ with $\delta_{n+1}-\delta_n<\varepsilon$. We "redefine" $f$ on each of $E_n:=f^{-1}(\delta_n,\delta_{n+1}]$ for $n\ge 0$. Since $E_n$ can still be unbounded and wild, let's further partition it into $$E_{nk} = (f^{-1}(\delta_n,\delta_{n+1}])\cap [k,k+1).$$ Now we "linearize" $f$ on $E_{nk}$ by setting $g|_{E_{nk}}$ to be a linear segment joining the points $(k,\delta_n')$ and $(k+1,\delta_n'')$ for $x\in E_{nk}$, where $\delta_n<\delta_n'<\delta_n''<\delta_{n+1}$.

The intervalwise linearization on $E_{nk},k=1,2,\dots$ shows that $g|_{E_{nk}}$ is injective for each $k$ and thus $g|_{E_{n}}^{-1}(\{r\}) = \cupp_{k\in \Z} g|_{E_{nk}}^{-1}(\{r\})$ is countable for every $n$, and  hence $g|_{E_{n}}^{-1}(\{r\})$ has measure zero, so does $g^{-1}(\{r\})$.

Finally we verify that $\|f-g\|_\infty<\varepsilon$. For every $x\in \R$, $x\in f^{-1}(\delta_n,\delta_{n+1}]$ for some $n\ge 0$, and hence $x\in E_{nk}$ for some $k$, but then both $f(x),g(x)\in (\delta_n,\delta_{n+1}]$, so $|f(x)-g(x)|<\varepsilon$.$\qed$