UMD Analysis Qualifying Exam/Aug08 Real

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Problem 1[edit]

Suppose that  \{f_n \} \!\, is a sequence of absolutely continuous functions defined on [0,1] \!\, such that f_n(0)=0 \!\, for every  n\!\, and


\sum_{n=1}^\infty  \int_0^1 |f_n^\prime(x)|dx < + \infty \!\,


for every x \in [0,1] \!\,. Prove:


  • the series \sum_{n=1}^\infty f_n(x) \!\, converges for each x \in [0,1] \!\, pointwise to a function  f\!\,


  • the function f \!\, is absolutely continuous on [0,1] \!\,


  • f^\prime(x)=\sum_{n=1}^\infty f_n^\prime(x) \quad a.e. \,\, x \in[0,1] \!\,


Solution 1a[edit]

Absolutely Continuous <==> Indefinite Integral[edit]

f_n(x) \!\, is absolutely continuous if and only if f_n(x) \!\, can be written as an indefinite integral i.e. for all x \in [0,1] \!\,



\begin{align}
f_n(x)-\underbrace{f_n(0)}_0 &= \int_0^{x} f_n^\prime(t) dt \\
                     f_n(x)  &= \int_0^{x} f_n^\prime(t) dt 
\end{align}

Apply Inequalities,Sum over n, and Use Hypothesis[edit]

Let x_0 \in [0,1] \!\, be given. Then,



\begin{align}
f_n(x_0) &= \int_0^{x_0} f_n^\prime(t) dt \\
         &\leq \int_0^{x_0} |f_n^\prime(t)| dt \\
         &\leq \int_0^1 |f_n^\prime(t)| dt
\end{align}


Hence


f_n(x_0) \leq \int_0^1 |f_n^\prime(t)|dt  \!\,


Summing both sides of the inequality over n \!\, and applying the hypothesis yields pointwise convergence of the series f_n \!\,,



\sum_{n=1}^\infty f_n(x_0) \leq \sum_{n=1}^\infty \int_0^1 |f_n^\prime(t)|dt  < +\infty
\!\,

Solution 1b[edit]

Absolutely continuous <==> Indefinite Integral[edit]

Let f(x)=\sum_{n=1}^\infty f_n(x) \!\,.


We want to show:


f(x)=\int_0^x \sum_{n=1}^\infty f^\prime_n (t)dt \!\,


Rewrite f(x) and Apply Lebesgue Dominated Convergence Theorem[edit]


\begin{align}
f(x)    &= \sum_{n=1}^\infty f_n(x)  \\
        &= \sum_{n=1}^\infty \int_0^x f_n^\prime(t) dt \quad \mbox{ (since } f_n \mbox{ is absolutely continuous)} \\
        &=\lim_{n \rightarrow \infty} \sum_{k=1}^n \int_0^x f_k^\prime(t) dt \\
        &=\lim_{n \rightarrow \infty} \int_0^x \sum_{k=1}^n f_k^\prime(t) dt \\
        &=\int_0^x \lim_{n \rightarrow \infty} \sum_{k=1}^n f_k^\prime(t) dt \quad \mbox{ (by LDCT)} \\
        &=\int_0^x \sum_{n=1}^\infty f_n^\prime(t) dt 

\end{align}

Justification for Lebesgue Dominated Convergence Theorem[edit]


\begin{align}
| \sum_{k=1}^n f^\prime_k(t) | &\leq \sum_{k=1}^n |f^\prime_k(t)| \\
                               &\leq \underbrace{\sum_{n=1}^\infty |f_n^\prime(t)|}_{g(t) \mbox{ dominating function} }\\
\\
\\
\int_0^1 \sum_{n=1}^\infty |f^\prime_n(x)|dx &= \sum_{n=1}^\infty  \int_0^1 
|f^\prime_n(x)|dx \quad \mbox{ (by Tonelli Theorem)}  \\
                        &< \infty \quad \mbox{ (by hypothesis) }
\end{align}


Therefore g(t) \!\, is integrable


The above inequality also implies \sum_{n=1}^\infty |f^\prime_n(x)| < \infty \!\, a.e on  [0,1] \!\,. Therefore,


| \sum_{k=1}^n f^\prime_k(t) | \rightarrow | \sum_{k=1}^\infty f^\prime_k(t) | \!\,


a.e on  [0,1] \!\, to a finite value.

Solution 1c[edit]

Since  f(x)=\int_{0}^{x}\sum_{n=1}^{\infty} f'_{n}(t)dt   \!\,,   by the Fundamental Theorem of Calculus

 f'(x)=\sum_{n=1}^{\infty} f'_{n}(x) \!\,   a.e. x \in [0,1] \!\,

Problem 3[edit]

Suppose that \{ f_n\} \!\, is a sequence of nonnegative integrable functions such that f_n \rightarrow f \!\, a.e., with f \!\, integrable, and \int_R f_n \rightarrow \int_R f  \!\,. Prove that


\int_R |f_n -f | \rightarrow 0 \!\,

Solution 3[edit]

Check Criteria for Lebesgue Dominated Convergence Theorem[edit]

Define   \hat{f}_n = |f-f_n|  \!\,,   g_n= f+f_n    \!\,.

g_n dominates hat{f}_n[edit]

Since   f_n     \!\, is positive, then so is   f  \!\, , i.e.,  |f_n|=f_n     \!\, and  |f|=f     \!\,. Hence,

|\hat{f}_n| = |f-f_n| \leq |f|+|f_n| = f + f_n = g_n     \!\,

g_n converges to g a.e.[edit]

Let g = 2f      \!\,. Since  f_n \rightarrow f    \!\,, then

 f+f_n \rightarrow 2f     \!\, , i.e.,

 g_n  \rightarrow g   \!\,.

integral of g_n converges to integral of g =[edit]


\begin{align}
\int g &= \int(f+f)\\
       &=2 \int f \\
\\
\\
\lim_{n \rightarrow \infty} \int g_n &= \lim_{n \rightarrow \infty} \int (f+f_n) \\
     &= \int f + \lim_{n\rightarrow \infty} \int f_n \\
     &= \int f + \int f \mbox{ (from hypothesis) } \\
     &= 2 \int f
\end{align}


Hence,


 \int g = \lim_{n \rightarrow \infty} \int g_n \!\,

hat{f_n} converges to hat{f} a.e.[edit]

Note that f_n \rightarrow f \!\, is equivalent to


|f_n -f | \rightarrow 0  \!\,


i.e.


 \hat{f_n} \rightarrow 0 = \hat{f} \!\,

Apply LDCT[edit]

Since the criteria of the LDCT are fulfilled, we have that

 \lim_n \int \hat{f_n} = \int \hat{f} = 0      \!\, , i.e.,

 \lim_n \int |f-f_n| =0     \!\,

Problem 5a[edit]

Show that if f \!\, is absolutely continuous on [0,1] \!\, and p>1 \!\,, then  |f|^p\!\, is absolutely continuous on [0,1] \!\,

Solution 5a[edit]

Show that g(x)=|x|^p is Lipschitz[edit]

Consider some interval I=[\alpha,\beta] \!\, and let x \!\, and y \!\, be two points in the interval I \!\,.


Also let K=\|g(x)\|_\infty  \!\, for all x \in I \!\,



\begin{align}
||x|^p-|y|^p| &=||x|-|y|| (|x|^{p-1}+|x|^{p-2}|y|+\ldots+|x||y|^{p-2}+|y|^{p-1}) \\
              &\leq |K^{p-1}+K^{p-2}K+\ldots+KK^{p-2}+K^{p-1}| ||x|-|y||\\
              &=\underbrace{pK^{p-1}}_M ||x|-|y|| \\
              &\leq M |x-y| 
\end{align}


Therefore g(x) \!\, is Lipschitz in the interval I \!\,

Apply definitions to g(f(x))[edit]

Since f(x) \!\, is absolutely continuous on [0,1] \!\,, given \epsilon >0 \!\,, there exists \delta >0 \!\, such that if \{(x_i,x_i^'\} \!\, is a finite collection of nonoverlapping intervals of [0,1] \!\, such that


\sum_{i=1}^n |x_i^'-x_i| < \delta \!\,


then


\sum_{i=1}^n |f(x_i^')-f(x_i)| < \epsilon \!\,


Consider g \circ f(x) = |f(x)|^p \!\,. Since g \!\, is Lipschitz



\begin{align}
\sum_{i=1}^n | g(f(x_i^'))-g(f(x_i)) | &\leq \sum_{i=1}^n M|f(x_i^')-f(x_i)| \\
                                       &=M \underbrace{\sum_{i=1}^n |f(x_i^')-f(x_i)|}_{< \epsilon} \\
                                       &< M\epsilon
\end{align}


Therefore g \circ f(x) = |f(x)|^p \!\, is absolutely continuous.

Problem 5b[edit]

Let 0<p<1 \!\,. Give an example of an absolutely continuous function f \!\, on [0,1] \!\, such that |f|^p \!\, is not absolutely continuous

Solution 5b[edit]

f(x)= x^4sin^2(\frac{1}{x^2}) is Lipschitz (and then AC)[edit]

Consider   f(x)= x^4\sin^2(\frac{1}{x^2})   \!\,. The derivate of f is given by

  f'(x)= 4x^3\sin^2(\frac{1}{x^2}) -2 x \sin(\frac{2}{x^2})              \!\,.

The derivative is bounded (in fact, on any finite interval), so   f  \!\, is Lipschitz.

Hence, f is AC

|f|^{1/2} is not of bounded variation (and then is not AC)[edit]

  |f(x)|^{1/2} = x^2\left|\sin \left(\frac{1}{ x^2}\right)\right|           \!\,

Consider the partition   \left\{\sqrt{\frac{2}{n \pi}}\right\}    \!\,. Then,

 \left|f\left ( \sqrt{\frac{2}{n \pi}}\right )\right|^{1/2} =  \frac{2}{n \pi} \left| \sin \left(\frac{n\pi}{2} \right) \right| \!\,

Then, T(f) goes to    \infty            \!\, as                n \!\, goes to      \infty          \!\,.

Then,       |f|^{1/2}         \!\, is not of bounded variation and then is not AC