# UMD Analysis Qualifying Exam/Jan07 Real

## Problem 1

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

### Solution 1

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

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

#### Case 1: ${\displaystyle \int _{1}^{\infty }f(x)<\infty }$

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

#### Case 1: ${\displaystyle \int _{1}^{\infty }f(x)=\infty }$

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

## Problem 3

 Let ${\displaystyle f\in L^{p}[0,1],g\in L^{q}[0,1],h\in L^{r}[0,1]}$ , where ${\displaystyle 1\leq p,q,r\leq \infty ,1/p+1/q+1/r=1}$ . Prove that ${\displaystyle fgh\in L^{1}[0,1]}$ , and ${\displaystyle ||fgh||_{1}\leq ||f||_{p}||g||_{q}||h||_{r}}$ .

### Solution 3

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

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

## Problem 5

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

### Solution 5

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

Then ${\displaystyle ||\mathrm {X} _{E_{n}}-f||_{1}=\int _{\mathbb {R} \setminus A}|f_{n}-f|\,dm+\int _{A}|f_{n}-f|\,dm=0+\int _{A}|f_{n}-f|\,dm}$

On ${\displaystyle A}$ , ${\displaystyle |f_{n}-f|}$  is a strictly positive function. Then for any ${\displaystyle \epsilon >0}$  sufficiently small there exists some ${\displaystyle A_{\epsilon }\subseteq A}$  with ${\displaystyle m(A_{\epsilon })>m(A)-\epsilon }$  such that ${\displaystyle |f_{n}-f|\geq C_{\epsilon }}$  on ${\displaystyle A_{\epsilon }}$  for some positive constant ${\displaystyle C_{\epsilon }}$ . Then ${\displaystyle ||\mathrm {X} _{E_{n}}-f||_{1}\geq C_{\epsilon }(m(A)-\epsilon )}$ .

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