Exercise: Suppose f:\mathbb{R}\rightarrow\mathbb{R} is continuous on \mathbb{R}, and differentiable in \mathbb{R}\backslash\{a\}. Suppose further that \lim_{x\rightarrow a}f'(x) exists. Prove that f is differentiable at a, and f'(a)=\lim_{x\rightarrow a}f'(x).
Proof. Let \{a_n\} be a sequence of real numbers approaching a, but none of them are equal to a. We want to show that under the given assumption the sequence \frac{f(a_n)-f(a)}{a_n-a} has a limit. By the mean value theorem, for each a_n, there is a b_n sitting between a and a_n such that \frac{f(a_n)-f(a)}{a_n-a}=f'(b_n). As b_n is also converging to a and f'(x) is continuous at a. So the limit exists and it’s equal to \lim_{x\rightarrow a}f'(x).
Exercise: Suppose for some integer n\geq 2 that the derivatives f', \ldots, f^{(n)} exist and are continuous on an open interval I containing x_0 and that f'(x_0)=\ldots=f^{(n-1)}(x_0)=0, but f^{(n)}\neq 0. Use Taylor’s Theorem to prove:
- If n is even and f^{(n)}(x_0)<0 then f has a local maximum at x_0.
- If n is even and f^{(n)}(x_0)>0 then f has a local minimum at x_0.
- If n is odd then f has neither a local maximum nor a local minimum at x_0.
Proof. By the assumptions, f(x)=f(x_0)+(a_n+h(x))(x-x_0)^n where a_n=\frac{f^{(n)}(x_0)}{n!} and \lim_{x\rightarrow x_0}h(x)=0.
- If a_n>0, as \lim_{x\rightarrow x_0}h(x)=0, there is a neighborhood of x_0, denoted as I_0, such that a_n+h(x)<0 for every x\in I_0. Since n is even, f(x)-f(x_0)=(a_n+h(x))(x-x_0)^n\leq 0 for every x\in I_0. And the equality holds if and only if x=x_0. So f achieves a local miximum at x_0.
- The same reasoning as above but with all the inequalities reversed.
- Suppose a_n>0. The same argument tells us that for every x\in I_0, f(x)-f(x_0)=(a_n+h(x))(x-x_0)^n has the same sign as x-x_0 if n is odd. Thus f has no local extremal at x_0. Suppose a_n<0, one can argue f(x)-f(x_0)=(a_n+h(x))(x-x_0)^n has the opposite sign of x-x_0. Thus the same.
Exercise:
- Suppose f(x)=c for all x\in[0,1]. Prove that f is Riemann integrable and \int_0^1 f(x)dx=c.
- Suppose f(x)=x for all x\in[0,1]. Prove that f is Riemann integrable and \int_0^1f(x)dx=\frac{1}{2}.
- Suppose f:[0,1]\rightarrow\mathbb{R}, f(x)=0 for all x\in[0,1] except that f(x)=1 for x=\frac{1}{2}. Prove that f is Riemann integrable and \int_0^1f(x)dx=0.
Proof.
- Whichever partition you choose, the Riemann sum is always c. So the integral is c.
- Let a partition P be 0 = x_0 < x_1 < \ldots < x_n = 1) with all gaps less than \delta. The the lower Riemann sum is x_0(x_1-x_0)+x_1(x_2-x_1)+\ldots+x_{n-1}(x_n-x_{n-1}), while the upper one is x_1(x_1-x_0)+x_2(x_2-x_1)+\ldots+x_{n}(x_n-x_{n-1}). The difference between the two sums is (x_1-x_0)^2+(x_2-x_1)^2+\ldots+(x_n-x_{n-1})^2, which is less than \delta((x_1-x_0)+(x_2-x_1)+\ldots+(x_n-x_{n-1}))=\delta. So f is Riemann integrable. To calculate to value of the integral, just look at the average of the lower sum and the upper sum, which turns out to be 1/2.
- Same argument as the previous one, but upper one is less than \delta, while the lower equals 0. Hence the integral is 0.
Exercise: Suppose f:[0,1]\rightarrow\mathbb{R} is continuous, f(x)\geq 0 for all x\in[0,1], and \int_0^1f(x)dx=0. Prove that f(x)=0 for all x\in[0,1].
Proof. (Sketch) Suppose for contradiction that f(x_0)>0 for some x_0\in[0,1]. As f is continuos, there is a non-empty interval (a,b)\subset[0,1] on which f is bigger than h=f(x_0)/2>0. Thus the integral is greater than or equal to h(a-b) which is positive. Contradiction.