UMD Analysis Qualifying Exam/Aug12 Real

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,

$$\lim_{n\rightarrow \infty} \int_0^\infty \frac{\sin(x/n)}{(1+x/n)^n} = \int_0^\infty \lim_{n\rightarrow \infty} \frac{\sin(x/n)}{(1+x/n)^n} = \int_0^\infty 0 = 0$$

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 $$g(x)$$.

Solution 5
(i) Fix epsilon greater than zero. Then, consider the sets Sn={x in [0,1] : n-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.