Calculus TA

Problem A real-valued function $f$ defined in $(a,b)$ is said to be convex if $$ f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda) f(y) $$ whenever $a<x<b,a<y<b,0<\lambda<1$. Prove that every convex function is continuous. Prove that every increasing convex function of a convex function is convex. (For example, if $f$ is convex, so is $e^f$.) If $f$ is convex in $(a,b)$ and if $a<s<t<u<b$, show that $$ \frac{f(t) - f(s)}{t-s} \leq \frac{f(u) - f(s)}{u-s} \leq \frac{f(u) - f(t)}{u-t}. $$ Answer (1) Let $f$ be a convex function in $(a,b)$ and $a<s<t<u<b$. Then note that $$ t = \frac{t-s}{u-s}u + (1-\frac{t-s}{u-s})s. $$ Then $$ f(t) \leq \frac{t-s}{u-s}f(u) + (1-\frac{t-s}{u-s})f(s). $$ From the above inequality, we can deduce the following two inequalities. $$ \frac{f(t) - f(s)}{t-s} \leq \frac{f(u) - f(s)}{u-s}, $$ $$ \frac{f(u) - f(s)}{u-s} \leq \frac{f(u) - f(t)}{u-t}. $$ (2) Fix $x \in (a,b)$ and $\epsilon >0$. For $y \in (a,b)$

Problem If $\sum a_n$ converges, and if $\{b_n\}$ is monotonic and bounded, prove that $\sum a_nb_n$ converges. Answer Since $b_n$ is monotonic and bounded, $b_n \rightarrow b$. If $b_n$ is increasing, then $\sum a_n (b-b_n)$ converges by Theorem 3.42. Then $\sum a_nb_n = \sum a_n b - \sum a_n (b - b_n)$ also converges. If $b_n$ is decreasing, then $\sum a_n (b_n-b)$ converges by Theorem 3.42. Then $\sum a_nb_n = \sum a_n b + \sum a_n (b_n - b)$ also converges. 

Problem Suppose (a) $f$ is continuous for $x \geq 0$, (b) $f'(x)$ exists for $x > 0$, (c) $f(0)=0$, (d) $f'$ is monotonically increasing. Put $$ g(x) = \frac{f(x)}{x}\;\;(x>0) $$ and prove that $g$ is monotonically increasing. Answer For $x>0$, $g'(x) = \frac{xf'(x) - f(x)}{x^2}$. Note that there exists $c \in (0,x)$ such that $g(x) = \frac{f(x)}{x} = \frac{f(x) - f(0)}{x- 0} = f'(c)$  $\Rightarrow$  $xf'(x) > xf'(c) = f(x)$  $\Rightarrow$  $g'(x)>0$.

Problem Suppose $g$ is a real function on $\mathbb{R}^1$, with bounded derivative (say $|g'| \leq M$). Fix $\epsilon >0$, and define $f(x) = x + \epsilon g(x)$. Prove that $f$ is one-to-one if $\epsilon$ is small enough. (A set of admissible values of $\epsilon$ can be determined which depends only on $M$.) Answer It is enough to show that $f'>0$ for a small $\epsilon>0$. Note that $$f'(x) = 1+ \epsilon g'(x) \geq 1 - \epsilon M.$$ Hence, if we choose $\epsilon < 1/M$, $f$ is one-to-one.

Problem Let $f$ and $g$ be continuous mappings of a metric space $X$ into a metric space $Y$, and let $E$ be a dense subset of $X$. Prove that $f(E)$ is dense in $f(X)$. If $g(p)=f(p)$ for all $p \in E$, prove that $g(p)=f(p)$ for all $p \in X$. (In other words, a continuous mapping is determined by its values on a dense subset of its domain.) Answer For $y \in f(X)$, there exists $x \in X$ such that $y=f(x)$. For $\epsilon > 0$, there exists $\delta > 0$ such that $d_X(x,x') < \delta$  $\Rightarrow$  $d_Y(f(x),f(x'))<\epsilon$. Since $E$ is dense in $X$, there exists $z \in E$ such that $z \in N_\delta(x)$, which implies $f(x) \in N_\epsilon(f(z))$. This means $f(E)$ is dense in $f(X)$. For $p \in X$ and $\epsilon >0$, there exists $\delta > 0$ such that $d_X(x,x') < \delta$  $\Rightarrow$  $d_Y(f(x),f(x'))<\epsilon/2, d_Y(g(x),g(x'))<\epsilon/2$. For $q \in E$ satisfying $q \in N_\delta(p)$, $d_Y(g(p),f(p)) \leq d_Y(g(p),g(q)) + d_Y(g(q)

Problem If $f$ is a continuous mapping of a metric space $X$ into a metric space $Y$, prove that $$f(\overline{E}) \subset \overline{f(E)}$$ for every set $E \subset X$. ($\overline{E}$ denotes the closure of $E$.) Show, by an example, that $f(\overline{E})$ can be a proper subset of $\overline{f(E)}$. Answer For $y \in f(\overline{E})$, there exists $x \in \overline{E}$ such that $y=f(x)$. For $\epsilon > 0$, there exists $\delta > 0$ such that $d_X(x,x')<\delta$  $\Rightarrow$  $d_Y(f(x),f(x'))<\epsilon$. Since $x \in \overline{E}$, there exists $z \in E$ such that $d_X(x,z) < \delta$, which implies $f(z) \in N_\epsilon(f(x))$. This means $y=f(x) \in \overline{f(E)}$.