UMD Analysis Qualifying Exam/Jan07 Real

Problem 1

Suppose that $f:[0,\mathrm{\infty })\to [0,\mathrm{\infty })$ is measurable and that $\underset{0}{\overset{1}{\int }}f(x)dx<\mathrm{\infty }$. Prove that $\underset{n\to \mathrm{\infty }}{\lim }\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{n}f(x)}{1+x^n}dx=\underset{0}{\overset{\mathrm{\infty }}{\int }}f(x)dx.$

Solution 1

Since $\underset{0}{\overset{1}{\int }}f(x)dx<\mathrm{\infty }$ then $f(x)<\mathrm{\infty }$ a.e. on [0,1]. This implies that $\frac{x^{n}f(x)}{1+x^n}\to 0$ a.e. on [0,1] since $\frac{x^n}{1+x^n}\to 0$ on (0,1). Then by Lebesgue Dominated Convergence, we have $\underset{n\to \mathrm{\infty }}{\lim }\underset{0}{\overset{1}{\int }}\frac{x^{n}f(x)}{1+x^n}dx=\underset{0}{\overset{1}{\int }}f(x)dx.$

Now to handle the interval $(1,\mathrm{\infty })$:

Case 1: $\underset{1}{\overset{\mathrm{\infty }}{\int }}f(x)<\mathrm{\infty }$

For every $x\in [1,\mathrm{\infty })$ we have $\frac{x^n}{1+x^n}<1$ and increases monotonically to 1. So by the same argument as above, Lebesgue Dominated convergence gives us $\underset{n\to \mathrm{\infty }}{\lim }\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{x^{n}f(x)}{1+x^n}dx=\underset{1}{\overset{\mathrm{\infty }}{\int }}f(x)dx.$ and we're done.

Case 1: $\underset{1}{\overset{\mathrm{\infty }}{\int }}f(x)=\mathrm{\infty }$

Notice that for every $x\in (1,\mathrm{\infty })$ we have $\frac{x^{n}f(x)}{1+x^n}>\frac{f(x)}{2}$, the right-hand-side must necessarily integrate to infinity. So by monotonicity of the integral we have that $\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{x^{n}f(x)}{1+x^n}=\mathrm{\infty }$ for all $n$. This gives $\underset{n\to \mathrm{\infty }}{\lim }\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{x^{n}f(x)}{1+x^n}dx=\underset{0}{\overset{\mathrm{\infty }}{\int }}f(x)dx=\mathrm{\infty }$ as desired.

Problem 3

Let $f\in L^p[0,1],g\in L^q[0,1],h\in L^r[0,1]$, where $1\le p,q,r\le \mathrm{\infty },1/p+1/q+1/r=1$. Prove that $fgh\in L^1[0,1]$, and $||fgh|{|}_1\le ||f|{|}_p||g|{|}_q||h|{|}_r$.

Solution 3

Dividing $f$ by $||f|{|}_p$, we can assume without loss of generality that $||f|{|}_p=1$ (similarly for $g,h$ with their appropriate norms). Thus we want to show that $||fgh|{|}_1\le 1$. The proof hinges on Young's Inequality which tells us that

$\int |fgh|\le \int \frac{|f{|}^p}{p}+\frac{|g{|}^q}{q}+\frac{|h{|}^r}{r}=\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=1.$

Problem 5

Suppose $E_n$ are measurable sets, and there is an integrable function $f\in L^1(ℝ)$ such that $\underset{n\to \mathrm{\infty }}{\lim }||{\mathrm{X}}_{E_n}-f|{|}_1=0$. Prove that there is a measureable set $E$ such that $f={\mathrm{X}}_E$.

Solution 5

We claim that $f$ can only take on the values 0 and 1. To see this, suppose the contrary, suppose $f$ differs from 0 or 1 on a set $A$. We can exclude the case $m(A)=0$ because otherwise we can modify $f$ on a null set to equal an indicator function without affecting the integral.

Then $||{\mathrm{X}}_{E_n}-f|{|}_1=\underset{ℝ\setminus A}{\int }|f_n-f|dm+\underset{A}{\int }|f_n-f|dm=0+\underset{A}{\int }|f_n-f|dm$

On $A$, $|f_n-f|$ is a strictly positive function. Then for any $ϵ>0$ sufficiently small there exists some $A_{ϵ}\subseteq A$ with $m(A_{ϵ})>m(A)-ϵ$ such that $|f_n-f|\ge C_{ϵ}$ on $A_{ϵ}$ for some positive constant $C_{ϵ}$. Then $||{\mathrm{X}}_{E_n}-f|{|}_1\ge C_{ϵ}(m(A)-ϵ)$.

Thus we have shown that we can obtain a positive lower bound for $||{\mathrm{X}}_{E_n}-f|{|}_1$ completely independent of the choice of $n$. This contradicts $\underset{n\to \mathrm{\infty }}{\lim }||{\mathrm{X}}_{E_n}-f|{|}_1=0$. Hence $f$ can only assume values 0 and 1 almost everywhere. Since $f\in L^1(ℝ)$, then it is certainly measurable. Hence $E:=f^{-1}(1)$ is measurable. And we're done.

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