UMD Analysis Qualifying Exam/Aug12 Real

Problem 1

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

Solution 1

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

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

Therefore,

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

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

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

Therefore, by the LDCT,

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

Problem 3

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

Solution 3

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

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

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

Therefore,

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

The same argument gives a lower bound, giving us altogether

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

Problem 5

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

Solution 5

(i) Fix epsilon greater than zero. Then, consider the sets S_n={x in [0,1] : n-1<=f(x)<n}. The partial sums of the integrals of f over S_n 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 A^c 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.