UMD Analysis Qualifying Exam/Aug12 Real

Problem 1

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

Solution 1

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

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

Therefore,

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

and this function is in $L_1([0,\mathrm{\infty }])$, with

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

Therefore, by the LDCT,

$\underset{n\to \mathrm{\infty }}{\lim }\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{\sin (x/n)}{(1+x/n{)}^n}=\underset{0}{\overset{\mathrm{\infty }}{\int }}\underset{n\to \mathrm{\infty }}{\lim }\frac{\sin (x/n)}{(1+x/n{)}^n}=\underset{0}{\overset{\mathrm{\infty }}{\int }}0=0$

Problem 3

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

Solution 3

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

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

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

Therefore,

$\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}=\underset{h\to 0}{\lim }\frac{\underset{x}{\overset{x+h}{\int }}g(t)dt}{h}=\underset{\begin{array}{c}h\to 0\\ h<\delta \end{array}}{\lim }\frac{\underset{x}{\overset{x+h}{\int }}g(t)dt}{h}<\underset{\begin{array}{c}h\to 0\\ h<\delta \end{array}}{\lim }\frac{\underset{x}{\overset{x+h}{\int }}g(x)+ϵdt}{h}=g(x)+ϵ$.

The same argument gives a lower bound, giving us altogether

$g(x)-ϵ<\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}<g(x)+ϵ$. 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 $ϵ>0$, there exists a $c>0$ such that $\underset{\{x\in [0,1]:f(x)\ge c\}}{\int }fdm<ϵ.$ (ii) Prove that, for each $ϵ>0$, there is a $\delta >0$ such that for each measurable subset $E\subseteq [0,1]$: if $m(E)<\delta$, then $\underset{E}{\int }fdm<ϵ.$

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.Template:BookCat