# UMD Analysis Qualifying Exam/Aug12 Real

## Problem 1

 Compute the following limit. Justify your answer. ${\displaystyle \lim _{n\to \infty }\int _{0}^{\infty }\left(1+{\frac {x}{n}}\right)^{-n}\sin \left({\frac {x}{n}}\right)\,dx.}$

### Solution 1

We will use the dominated convergence theorem. First, note that for ${\displaystyle x\geq 0}$ and ${\displaystyle n\geq 2}$,

${\displaystyle (1+x/n)^{n}=1+x+{\frac {n-1}{2n}}x^{2}+...\geq 1+x+{\frac {n-1}{2n}}x^{2}\geq 1+x+{\frac {1}{4}}x^{2}=(1+x/2)^{2}.}$

Therefore,

${\displaystyle {\frac {\sin(x/n)}{(1+x/n)^{n}}}\leq {\frac {1}{(1+x/2)^{2}}}}$

and this function is in ${\displaystyle L_{1}([0,\infty ])}$, with

${\displaystyle \int _{0}^{\infty }{\frac {1}{(1+x/2)^{2}}}=2}$.

Therefore, by the LDCT,

${\displaystyle \lim _{n\rightarrow \infty }\int _{0}^{\infty }{\frac {\sin(x/n)}{(1+x/n)^{n}}}=\int _{0}^{\infty }\lim _{n\rightarrow \infty }{\frac {\sin(x/n)}{(1+x/n)^{n}}}=\int _{0}^{\infty }0=0}$

## Problem 3

 Assume ${\displaystyle f}$ is absolutely continuous on an interval ${\displaystyle [a,b]}$ and there is a continuous function ${\displaystyle g}$ such that ${\displaystyle f'=g}$ a.e. Show that ${\displaystyle f}$ is differentiable at every ${\displaystyle x\in [a,b]}$ and that ${\displaystyle f'(x)=g(x)}$ everywhere on ${\displaystyle [a,b]}$.

### Solution 3

If ${\displaystyle f'(x)}$ exists, then by definition, ${\displaystyle f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}}$. So we need to show that this limit both exists and is equal to ${\displaystyle g(x)}$.

Then by the absolute continuity of ${\displaystyle f}$, ${\displaystyle \lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\to 0}{\frac {\int _{x}^{x+h}g(t)\,dt}{h}}}$.

Since, ${\displaystyle g(x)}$ is continuous, then for any ${\displaystyle \epsilon >0}$ there exists some ${\displaystyle \delta >0}$ such that for ${\displaystyle h<\delta }$,${\displaystyle |g(x+h)-g(x)|<\epsilon }$.

Therefore,

${\displaystyle \lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}=\lim _{h\to 0}{\frac {\int _{x}^{x+h}g(t)\,dt}{h}}=\lim _{\begin{array}{c}h\to 0\\h<\delta \end{array}}{\frac {\int _{x}^{x+h}g(t)\,dt}{h}}<\lim _{\begin{array}{c}h\to 0\\h<\delta \end{array}}{\frac {\int _{x}^{x+h}g(x)+\epsilon \,dt}{h}}=g(x)+\epsilon }$.

The same argument gives a lower bound, giving us altogether

${\displaystyle g(x)-\epsilon <\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}. Therefore, the limit exists (i.e. ${\displaystyle f}$ is differentiable) and the difference quotient goes to ${\displaystyle g(x)}$.

## Problem 5

 Let ${\displaystyle f}$ be a nonnegative Lebesgue integrable function on ${\displaystyle [0,1]}$. Denote by ${\displaystyle m}$ the Lebesgue measure on ${\displaystyle [0,1]}$. (i) Prove that, for each ${\displaystyle \epsilon >0}$, there exists a ${\displaystyle c>0}$ such that ${\displaystyle \int _{\{x\in [0,1]:f(x)\geq c\}}f\,dm<\epsilon .}$ (ii) Prove that, for each ${\displaystyle \epsilon >0}$, there is a ${\displaystyle \delta >0}$ such that for each measurable subset ${\displaystyle E\subseteq [0,1]}$: if ${\displaystyle m(E)<\delta }$, then ${\displaystyle \int _{E}f\,dm<\epsilon .}$

### Solution 5

(i) Fix epsilon greater than zero. Then, consider the sets Sn={x in [0,1] : n-1<=f(x)<n}. The partial sums of the integrals of f over Sn comprise a monotonically increasing sequence of real numbers, bounded by the finite integral of f over [0,1].

Hence, this sequence converges, and the tail of the sequence, which is the integral of f over the set {x in [0,1] : f(x)>= n}, must eventually be less than epsilon for some n.

(ii) Fix epsilon greater than zero. By part (i), there exists some constant c such that, given the set A={x in [0,1] : f(x)>=c}, the integral of f over A is less than epsilon/2. On the complement of A (in [0,1]), f is bounded above by c, and so any set of measure less than epsilon/2c will produce an integral whose value less than epsilon/2.

If m(A) is nonzero, take delta to be the minimum of m(A) and epsilon/2c. Given any measurable E in [0,1] with m(E) less than delta, the portion in A and the portion in Ac will each have integrals with values at most epsilon/2, so the integral of f over E has a value of at most epsilon.

If m(A)=0, then f is bounded almost everywhere on [0,1], and we simply take delta to be epsilon/c.