UMD Analysis Qualifying Exam/Jan07 Real

Problem 1

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

$\lim _{n\to \infty }\int _{0}^{\infty }{\frac {x^{n}f(x)}{1+x^{n}}}\,dx=\int _{0}^{\infty }f(x)\,dx.$

Solution 1

Since $\int _{0}^{1}f(x)\,dx<\infty$ then $f(x)<\infty$ 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 $\lim _{n\to \infty }\int _{0}^{1}{\frac {x^{n}f(x)}{1+x^{n}}}\,dx=\int _{0}^{1}f(x)\,dx.$

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

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

For every $x\in [1,\infty )$ 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 $\lim _{n\to \infty }\int _{1}^{\infty }{\frac {x^{n}f(x)}{1+x^{n}}}\,dx=\int _{1}^{\infty }f(x)\,dx.$ and we're done.

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

Notice that for every $x\in (1,\infty )$ 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 $\int _{1}^{\infty }{\frac {x^{n}f(x)}{1+x^{n}}}=\infty$ for all $n$ . This gives $\lim _{n\to \infty }\int _{0}^{\infty }{\frac {x^{n}f(x)}{1+x^{n}}}\,dx=\int _{0}^{\infty }f(x)\,dx=\infty$ as desired.

Problem 3

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

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 $||fgh||_{1}\leq 1$ . The proof hinges on Young's Inequality which tells us that

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

Problem 5

Suppose $E_{n}$ are measurable sets, and there is an integrable function $f\in L^{1}(\mathbb {R} )$ such that $\lim _{n\to \infty }||\mathrm {X} _{E_{n}}-f||_{1}=0$ . 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 $||\mathrm {X} _{E_{n}}-f||_{1}=\int _{\mathbb {R} \setminus A}|f_{n}-f|\,dm+\int _{A}|f_{n}-f|\,dm=0+\int _{A}|f_{n}-f|\,dm$

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

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

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