Math 521 midterm 3

The third midterm will cover the material from chapters 4 and 5 on continuity and differentiability of Rudin’s book.

Practice problems

U N D E R       C O N S T R U C T I O N

Continuity in metric spaces

  1. Suppose $X$, $Y$, and $Z$ are metric spaces, and let $f:X\to Y$ and $g:Y\to Z$ be two functions. If $f$ is continuous at $p\in X$ and $g$ is continuous at $f(p)\in Y$, then show that $g\circ f:X\to Z$ is continuous at $p$.
  2. Prove that if $f:X\to Y$ is a continuous map, where $X$ and $Y$ are metric spaces, then $f^{-1}(E)$ is an open subset of $X$ whenever $E\subset Y$ is open. (In particular, begin by stating the definition of $f^{-1}(E)$.)
  3. True or False? If $f:X\to Y$ is a continuous map from one metric space to another, and if $E\subset X$ is open, then its image $f(E)$ is an open subset of $Y$.
  4. Let $E=\{p\in\R^2 : \|p\|\le 1, p\neq0\}$. Assume that $f:E\to\R$ is a uniformly continuous function on $E$. Show that $f$ is bounded.
  5. True or False?If $E\subset X$ is a closed subset, and if $f:X\to Y$ is a continuous mapping, then $f(E)$ is a closed subset of $Y$.
    Solution: Not true. Consider $E=X=\R$, $Y=\R$, and $f(x) = 1/(1+x^2)$. Or $E=(-\infty, 0]$, $X=Y=\R$ and $f(x) = e^x$.

Continuity in $\R$

  1. Let $C\subset\R^n$ be a closed set, and let $p\in\R^n$. Show that there is a point $q\in C$ that is closest to $p$, i.e. for every point $\tilde q\in C$ one has $|\tilde q-p| \geq |q-p|$.
  2. If $f:[a,b]\to \R$ is a continuous function, and if $f(a)\gt a$ and $f(b)\lt b$, then there is a $c\in(a,b)$ such that $f(c) = c$.
  3. If $J\subset \R$ is a closed interval, and $f:J\to\R$ is a continuous function with $f(x)\in J$ for all $x\in J$ (i.e., $f(J)\subset J$ for short) then show that there is a $c\in J$ with $f(c) = c$.
  4. Consider the function $f:(0,1)\to\R$ given by $f(x) = 1/x$.
    1. Is $f$ continuous on the interval $(0,1)$?
    2. Is $f$ uniformly continuous on the interval $(0,1)$?
    3. For which $a\gt 0$ (if any) is $f$ continuous on the interval $(a,1)$?
  5. True or False? If $f:\R\to\R$ is uniformly continuous on any bounded interval $[-M, M]$, then $f$ is uniformly continuous on $\R$.
    Solution: Not true. Consider the function $f(x) = x^2$. We know from class that it is not uniformly continuous on $\R$ (see below), but it is uniformly continuous on any compact interval $[-M,M]$, due to the general theorem that says A continuous function on a compact metric space is always uniformly continuous.

    Reason why $f(x)=x^2$ is not uniformly continuous. Consider this calculation: for any $x\gt 0$ define $\bar x=x+\frac1x$. Then \[ \bar x - x = \frac 1x, \] and \[ f(\bar x) - f(x) = \Bigl(x+\frac1x\Bigr)^2 - x^2 = 2 + \frac1{x^2} \gt 2. \] If $f$ were uniformly continuous, then we could choose $\varepsilon=1$, and there would be a $\delta\gt 0$ such that $|f(\bar x)- f(x)|\lt 1$ for all $x, \bar x$ with $|\bar x-x|\lt \delta$. But our calculation above shows that if we choose $x$ large enough (i.e. $x\gt \frac1\delta$), then we have $\bar x-x\lt \delta$, and $f(\bar x)-f(x)\gt 2\gt\varepsilon$.

  6. If $f:(0,1)\to\R$ is uniformly continuous, and if $x_n\in(0,1)$ is a sequence of real numbers with $x_n\to0$, then show that $\lim_{n\to\infty} f(x_n)$ exists (hint: show that $f(x_n)$ is a Cauchy sequence.)
    Solution: Let $\varepsilon\gt0$ be given. Then there is a $\delta\gt0$ such that $|f(\bar x)-f(x)|\lt \varepsilon$ holds for all $x, \bar x\in (0,1)$ with $|\bar x- x|\lt \delta$. Since $x_n\to0$ there is an $N\in\N$ such that $x_n \in (0, \delta)$ for all $n\ge N$.

    If $n, m\ge N$ then we have $x_n, x_m\in(0, \delta)$, and thus $|x_n-x_m|\lt \delta$. This then implies $|f(x_n)-f(x_m)|\lt \varepsilon$ for all $n, m\ge N$. Since this argument applies for any $\varepsilon\gt0$, we have shown that the sequence of real numbers $f(x_n)$ is a Cauchy sequence. It follows that $\lim_{n\to\infty}f(x_n)$ exists.

Derivatives

  1. If $f:J\to\R$ is a function on some open interval $J\subset \R$, and if $f$ is differentiable at $a\in J$, then show that $f$ is continuous at $a$.
  2. Suppose $f:[a,b]\to\R$ is a continuous function which is differentiable on $(a,b)$, and assume that $f'(x)\lt 0$ for all $x\in (a,b)$. Show that $f(a) \gt f(b)$.
  3. Suppose $f:\R\to\R$ is a continuous function which is twice differentiable. Assume there are $a\lt b\lt c$ such that $f(a) = f(b)= f(c)= 0$. Show that there is an $x\in(a, c)$ such that $f''(x)=0$.
    Solution: Apply Rolle’s theorem to $f$ on the interval $[a,b]$ to conclude that there is a $p\in(a,b)$ with $f'(p)=0$. Similarly, there is a $q\in(b, c)$ with $f'(q)=0$. Next, apply Rolle’s theorem to $f'$ on the interval $[p,q]$, and you find that there is an $r\in(p,q)$ such that $f''(r) = 0$.
  4. Suppose $f:\R\to\R$ is a continuous function which is twice differentiable, and suppose the graph of $f$ intersects a given straight line in three different points. Show that $f''(x)=0$ for at least one $x\in \R$.
    Solution: Assume the straight line has equation $y=mx+n$, and assume that the graph of $f$ intersects the line at $x=a$, $x=b$, and $x=c$, where we may assume that $a\lt b\lt c$. Apply the Mean Value Theorem theorem to $f$ on the interval $[a,b]$ to conclude that there is a $p\in(a,b)$ with $f'(p)=m$. Similarly, there is a $q\in(b, c)$ with $f'(q)=m$. Next, apply the MVTheorem to $f'$ on the interval $[p,q]$, and you find that there is an $r\in(p,q)$ such that $f''(r) = 0$.
  5. Let $f:[a,b]\to\R$ be a continuous function which is three times continuously differentiable, and assume its graph intersects the graph of the parabola $y=x^2$ in four different points. Show that there is an $x$ with $f^{(3)}(x) = 0$.
    Solution: Let the $x$ coordinates of the intersections points be $a\lt b\lt c\lt d$. Let the parabola be given by $y=P(x) = kx^2+\ell x+m$.

    Consider the function $g(x) = f(x) - P(x)$. This function vanishes at $a,b,c,d$, so by Rolle’s theorem there are $p,q,r$ with \[ a\lt p\lt b\lt q \lt c \lt r\lt d \] such that $g'(p)=g'(q)=g'(r)=0$. Apply Rolle’s theorem again, but now to $g'$, and you find that there are $s\in(p,q)$ and $t\in(q,r)$ with $g''(s) = g''(t) = 0$. Applying Rolle one more time then gives you a $u\in(s,t)$ with $g'''(u)=0$.

    This implies $f'''(u) = P'''(u)$. But $P(x)$ is a quadratic function, so its third derivative vanishes. Therefore $f'''(u) = 0$.

  6. Let $f:\R\to\R$ be a differentiable function. Assume that there are four numbers $a\lt b\lt c\lt d$ such that $f(b)\lt f(a)$ and $f(c)\lt f(d)$. Show that there is an $x\in (a, d)$ such that $f'(x)=0$.
    Solution: There are two theorems that we will use: a continuous function on a compact metric space attains its minimum and the derivative of a differentiable function on an interval vanishes at any interior minimum or maximum point.

    The interval $[a,d]$ is compact, and the function $f$ is continuous, so it attains its minimum at some $x\in[a,d]$. Since $f(b)\lt f(a)$ the minimum is not attained at $a$, so $x\in(a, d]$. Similarly, $f(c) \lt f(d)$, so the minimum is also not attained at $x=d$. Therefore $x\in(a,d)$ so the point $x$ at which the minimum is attained is an interior point, and we conclude that $f'(x)=0$.

  7. Let $f:\R\to\R$ be a twice differentiable function. Assume that there are four numbers $a\lt b\lt c\lt d$ such that $f(b)\lt f(a)$, $f(c)\gt f(b)$, and $f(d)\lt f(c)$. Show that there is an $x\in (a, d)$ such that $f''(x)=0$.
    Solution: The Mean Value Theorem applied to $f$ on each of the three intervals $(a,b)$, $(b,c)$, $(c, d)$, implies that there are $p\in(a,b)$, $q\in(b,c)$, and $r\in (c,d)$ with \[ f'(p)\lt 0, \quad f'(q)\gt0, \quad f'(r)\lt0. \] Since we are given that $f$ has two derivatives, we know that $f'$ is a differentiable function, and thus that $f'$ is continuous. The Intermediate Value Theorem then implies that there exist $s\in(p,q)$ and $t\in(q, r)$ with \[ f'(s) = 0, \quad f'(t)=0. \] Finally, we apply Rolle’s theorem to $f'$ on the interval $[s,t]$, and conclude that there is a $u\in(s,t)$ with $f''(u)=0$.