UMD Analysis Qualifying Exam/Aug07 Real

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,


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

Alternate Solution
Suppose $$\int_{-\infty}^\infty f' = c\neq 0$$ (without loss of generality, $$c>0$$). Then for small positive $$\epsilon$$, there exists some real $$M$$ such that for all $$m>M$$ we have $$c-\epsilon<\int_{-m}^m f' M$$.

Since $$f$$ is integrable, this means that for any small positive $$\delta$$, there exists an $$N$$ such that for all $$n>N$$, we have $$\int_{-\infty}^{-n} +\int_n^\infty f <\delta$$. But by the above estimate,

$$\int_{-\infty}^{-n} f +\int_n^\infty f >\int_{-\infty}^{-n} f + \int_{-\infty}^{-n} f +(c-\epsilon) = \int_{-\infty}^{-n} 2 f + \int_{-\infty}^{-n} (c-\epsilon) =\infty $$

This contradicts the integrability of $$f$$. Therefore, we must have $$c=0$$.

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,


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

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 Vitali Convergence Theorem then implies


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

Solution 5
$$ \begin{align} \int_R |f(x)|dx &= \int_{|x|>1} |xf(x| \cdot \frac{1}{|x|}dx + \int_{|x| <1} |f(x)| \cdot 1 dx \\ &\leq \left( \int_{|x| >1} |xf(x)|^2 dx \right)^{\frac{1}{2}} \cdot \left( \int_{|x|>1}\frac{1}{|x|^2}dx\right)^{\frac{1}{2}} + \left( \int_{|x|<1} |f(x)|^2 dx \right)^{\frac{1}{2}} \cdot \left(\int_{|x|<1} 1^2 dx \right)^{\frac{1}{2}} \end{align}        $$