UMD Analysis Qualifying Exam/Aug07 Real

Problem 1

Suppose that $f \!\,$ is a continuous real-valued function with domain $(-\infty,\infty) \!\,$ and that $f \!\,$ is absolutely continuous on every finite interval $[a,b] \!\,$. Prove: If $f \!\,$ and $f^{\prime} \!\,$ are both integrable on $(-\infty, \infty) \!\,$, then $\int_{-\infty}^{\infty} f^{\prime}=0 \!\,$

Solution 1

Since $f \!\,$ is absolutely continuous for all $[a,b] \subset R^1 \!\,$,

$\int_a^b f^\prime(x)dx= f(b)-f(a) \!\,$

Hence

$\int_{-\infty}^\infty f^\prime(x)dx= \lim_{a,b \rightarrow \infty} \int_a^b f^\prime(x)= \lim_{a,b \rightarrow \infty}[f(b)-f(a)] \!\,$

Since $f^\prime \!\,$ is integrable i.e. $f^\prime \in L^1 (-\infty, \infty) \!\,$, $\lim_{b \rightarrow \infty}f(b) \!\,$ and $\lim_{a \rightarrow \infty} f(a) \!\,$ exist.

Assume for the sake of contradiction that

$\lim_{b \rightarrow \infty} |f(b)|= \delta >0 \!\,$

Then there exists $M \!\,$ such that for all $x >M \!\,$

$|f(x)| > \frac{\delta}{2} \!\,$

since $f \!\,$ is continuous. (At some point, $|f| \!\,$ will either monotonically increase or decrease to $\delta \!\,$.) This implies

$\begin{align} \int_{-\infty}^{\infty} |f(x)|dx &\geq \int_M^\infty |f(x)|dx \\ &\geq \int_M^\infty \frac{\delta}{2}dx \\ &=\infty \end{align}$

which contradicts the hypothesis that $f \!\,$ is integrable i.e. $f \in L^1(-\infty, \infty) \!\,$. Hence,

$\lim_{b \rightarrow \infty} f(b) =0 \!\,$

Using the same reasoning as above,

$\lim_{a \rightarrow -\infty} f(a) =0 \!\,$

Hence,

$\int_{-\infty}^\infty f^\prime(x) dx = \lim_{b \rightarrow \infty} f(b)-\lim_{a \rightarrow -\infty} f(a) =0 \!\,$

Problem 3

Suppose that $\{f_n \} \!\,$ is a sequence of real valued measurable functions defined on the interval $[0,1] \!\,$ and suppose that $f_n(x) \rightarrow f(x) \!\,$ for almost every $x \in [0,1] \!\,$. Let $p >1 \!\,$ and $M >0 \!\,$ and suppose that $\|f_n\|_p \leq M \!\,$ for all $n \!\,$

Problem 3a

Prove that $\|f\|_p \leq M \!\,$.

Solution 3a

By definition of norm,

$\|f\|_p = \left(\int |f(x)|^p\,\mathrm dx\right)^{\frac1p} \!\,$

Since $\| f_n \|_p \leq M \!\,$,

$\|f_n\|_p^p \leq M^p \!\,$

By Fatou's Lemma,

$\begin{align} \|f\|_p^p &=\int_0^1 |f(x)|^p dx \\ &=\int_0^1 \underset{n}{\lim \inf} |f_n(x)|^p dx \\ &\leq \underset{n}{\lim \inf} \int_0^1 |f_n(x)|^p dx \\ &\leq M^p \end{align}$

which implies, by taking the $p\!\,$th root,

$\|f\|_p \leq M \!\,$

Problem 3b

Prove that $\|f-f_n\|_1 \rightarrow 0 \!\,$ as $n \rightarrow \infty \!\,$

Solution 3b

By Holder's Inequality, for all $A \subset [0,1] \!\,$ that are measurable,

$\begin{align} \int_A |(f(x)-f_n(x))\cdot 1|dx &\leq \left( \int_A |f(x)-f_n(x)|^p dx \right)^{\frac{1}{p}}\cdot \left(\int_A 1^qdx \right)^{\frac{1}{q}} \\ &\leq \left( \int_A |2f(x)|^p dx \right)^{\frac{1}{p}}\cdot \left(\int_A 1^qdx \right)^{\frac{1}{q}} \\ &\leq 2\left( \int_A |f(x)|^p dx \right)^{\frac{1}{p}}\cdot \left(\int_A 1^qdx \right)^{\frac{1}{q}} \\ &\leq 2M \cdot (m(A))^{\frac{1}{q}} \end{align}$

where $\frac1p + \frac1q=1 \!\,$

Hence, $|f(x)-f_n(x)| \leq 2M (m(A))^{\frac{1}{q}} \!\,$

The Dominated Convergence Theorem then implies

$\lim_{n \rightarrow \infty} \int_0^1 |f(x)-f_n(x)|dx = \int_0^1 \lim_{n \rightarrow \infty} |f(x)-f_n(x)|dx = 0 \!\,$

Problem 5

Suppose $f(x), xf(x) \in L^2(R) \!\,$. Prove that $f(x) \in L^1(R) \!\,$ and that $\|f\|_1 \leq \sqrt{2}( \|f\|_2+ \|x f\|_2) \!\,$

Solution 5

$\begin{align} \int_R |f(x)|dx &= \int_{|x|>1} |xf(x| \cdot \frac{1}{|x|}dx + \int_{|x| <1} |f(x)| \cdot 1 dx \\ &\leq \left( \int_{|x| >1} |xf(x)|^2 dx \right)^{\frac{1}{2}} \cdot \left( \int_{|x|>1}\frac{1}{|x|^2}dx\right)^{\frac{1}{2}} + \left( \int_{|x|<1} |f(x)|^2 dx \right)^{\frac{1}{2}} \cdot \left(\int_{|x|<1} 1^2 dx \right)^{\frac{1}{2}} \end{align}$