# Extra_Questions

DZG: this comes from some tex file that I found when studying for quals, so is definitely not my own content! I’ve just copied it here for extra practice.

# May 2016 Qual

## May 2016, 1

Consider the function $$f(x) = \frac{x}{1-x^2}$$, $$x \in (0,1)$$.

• By using the $$\epsilon-\delta$$ definition of the limit only, prove that $$f$$ is continuous on $$(0,1)$$. (Note: You are not allowed to trivialize the problem by using properties of limits).

• Is $$f$$ uniformly continuous on $$(0,1)$$? Justify your answer.

Fix $$x\in (0,1)$$ and let $$\epsilon > 0$$. Then we have \begin{align*}\left |f(x) - f(y) \right| = \left|\frac{x}{1-x^2} - \frac{y}{1-y^2}\right| = \left| \frac{x(1-y^2) - y(1-x^2)}{(1-x^2)(1-y^2)} \right| = \left| \frac{x-y}{(1-x)(1+x)(1-y)(1+y)} \right|.\end{align*} Now, choose $$\delta > 0$$ such that $$\delta < \min\{\frac{1}{2}(1-x)^2\epsilon, \frac{1}{2}(1-x)\}$$. When $$x - \delta < y < x + \delta$$, \begin{align*}\begin{aligned} |f(x) - f(y) | & = & \left| \frac{x-y}{(1-x)(1+x)(1-y)(1+y)} \right| \\ & \leq & \left| \frac{x-y}{(1-x)(1-y)} \right| \leq \left| \frac{x-y}{(1-x)(1-(x+ \frac{1}{2}(1 - x)))} \right| \\ & = & \left| \frac{x-y}{(1-x)(1-(x+ \frac{1}{2}(1 - x)))} \right| = \left| \frac{2}{(1-x)^2} \right||x-y| \\ & < & \epsilon. \end{aligned}\end{align*}

As our choice of $$x\in (0,1)$$ was arbitrary, we conclude that $$f$$ is continuous on all of $$(0,1)$$. ◻

Proof. We will show that the function $$f$$ is not uniformly continuous. Consider the sequence $$(x_n)_{n=1}^\infty$$ in $$(0,1)$$ defined by $$x_n = \frac{n}{n+1}$$. Observe that \begin{align*}f(x_n) = \frac{\frac{n}{n+1}}{1-\left(\frac{n}{n+1}\right)^2} = \frac{n(n+1)}{(n+1)^2 - n^2} = \frac{n(n+1)}{[(n+1)-n][(n+1)+n]} = \frac{n(n+1)}{2n+1}\end{align*} Written as $$x_n = 1 - \frac{1}{n+1}$$, one can more easily see that $$(x_n)_{n=1}^\infty$$ converges to $$1$$ in $$\mathbb{R}$$, hence is Cauchy in $$(0,1)$$. Now, let $$\delta > 0$$ and choose $$N\in \mathbb{N}$$ such that $$|x_n - x_m| < \delta$$ when $$n,m \geq N$$. For $$\epsilon < \frac{1}{8}$$ we have

\begin{align*}\begin{aligned} \left| f(x_n) - f(x_{n+1}) \right| &=& \left|\frac{n(n+1)}{2n+1} - \frac{(n+1)(n+2)}{2n+3} \right| = \left|\frac{n(n+1)(2n+3) - (n+1)(n+2)(2n+1)}{(2n+1)(2n+3)} \right| \\ &=& \left|\frac{(2n^3+5n^2+3n) - (2n^3+7n^2+7n+2)}{(2n+1)(2n+3)} \right| = \left|\frac{ 2n^2+4n+2 }{4n^2 + 8n + 3} \right| \\ &\geq& \left| \frac{2n^2}{ 16n^2 } \right| = \frac{1}{8}.\end{aligned}\end{align*} So for any $$\delta > 0$$, we see that there exists two points $$x_n, x_{n+1} \in (0,1)$$ such that $$|x_n - x_{n+1}| < \delta$$ when $$n$$ is sufficiently large but $$f(x_n) - f(x_{n+1}) | \not < \epsilon$$. Therefore $$f(x)$$ is not uniformly continuous. ◻

## (May 2016, 2)

Let $$\{a_k\}_{k=1}^\infty$$ be a bounded sequence of real numbers and $$E$$ given by: \begin{align*}E:= \bigg\{s \in \mathbb{R}\, \colon \, \text{ the set } \{k \in \mathbb{N}\, \colon \, a_k \geq s\} \text{ has at most finitely many elements}\bigg\}.\end{align*} Prove that $$\limsup_{k \to \infty} a_k = \inf E$$.

Proof. Let $$e \in E$$. As there are only finitely many $$a_k \geq s$$, there exists some $$N \in \mathbb{N}$$ such that $$a_k < e$$ for all $$k \geq N$$. Define $$T_k := \{a_k : k \geq n\}$$. It is clear that $$e$$ is thus an upper bound for $$T_N$$. So, \begin{align*}e \geq \sup T_N \geq \limsup a_k.\end{align*} Thus, $$\limsup a_k$$ is a lower bound for $$E$$, meaning $$\inf E \geq \limsup a_n$$.
Conversely, suppose $$k \in \mathbb{N}$$. \begin{align*}T_k = \{a_n : n \geq k \}.\end{align*} So, $$\sup T_k \geq a_n$$ for all $$a_n \in T_k$$. Then, $$\{a_k : a_k \geq \sup T_k\}$$ must be finite, so $$\{k \in \mathbb{N} : a_k \geq \sup T_k\}$$ is finite. So, $$\sup T_k \in E$$ for all $$k \in \mathbb{N}$$. Since $$\inf E$$ is a lower bound for $$E$$, $$\inf E \leq \sup T_k$$ for all $$k \in \mathbb{N}$$. Thus, \begin{align*}\inf E \leq \lim (\sup T_k) = \limsup a_k.\end{align*} We have both inequalities, therefore $$\limsup a_k = \inf E$$. ◻

## (May 2016, 3)

Assume $$(X,d)$$ is a compact metric space.

• Prove that $$X$$ is both complete and separable.

• Suppose $$\{x_k\}_{k=1}^\infty \subseteq X$$ is a sequence such that the series $$\sum_{k=1}^\infty d(x_k, x_{k+1})$$ converges. Prove that the sequence $$\{x_k\}_{k=1}^\infty$$ converges in $$X$$.

## (May 2016, 4)

Suppose that $$f \colon [0,2] \to \mathbb{R}$$ is continuous on $$[0,2]$$ , differentiable on $$(0,2)$$, and such that $$f(0) = f(2) = 0$$, $$f(c) = 1$$ for some $$c \in (0,2)$$. Prove that there exists $$x \in (0,2)$$ such that $$|f'(x)| >1.$$

Proof. We will consider three cases. First, suppose $$c<1$$. Then, by the mean value theorem, there exists $$x\in (0,c)$$ such that $$f'(x)(c-0)=f(c)-f(0)$$ so $$f'(x)=\frac{f(c)}{c}=\frac{1}{c}>1$$ since $$c<1$$. Similarly, if $$c>1$$ then by the mean value theorem there exits $$y\in (c,2)$$ such that \begin{align*}|f'(y)|=\left\lvert\frac{f(2)-f(c)}{2-c}\right\rvert=\left\lvert \frac{-f(c)}{2-c}\right\rvert=\left\lvert\frac{-1}{2-c}\right\rvert>1\end{align*} since $$1<c<2$$.

Now, suppose $$c=1$$. If there exists $$x\in (0,1)$$ such that $$x<f(x)$$ then by the mean value theorem on the interval $$(0,x)$$ there exists $$s\in (0,x)$$ such that $$f'(s)=\frac{f(x)}{x}>1$$ since $$f(x)>x$$. Likewise, if there exists $$x\in (0,1)$$ such that $$x>f(x)$$ then the mean value theorem on $$(x,1)$$ gives a point $$t\in (x,1)$$ such that $$\left\lvert f'(t)\right\rvert=\left\lvert \frac{f(1)-f(x)}{1-x}\right\rvert=\left\lvert\frac{1-f(x)}{1-x}\right\rvert>1$$ since $$x>f(x)$$. So, on $$(0,1)$$, if the proposition does not hold then $$f(x)=x$$. Similarly, if there exists $$x\in (1,2)$$ such that $$f(x)>2-x$$ then the mean value theorem yields a point $$u\in (x,2)$$ such that $$|f'(u)|=\left\lvert \frac{f(2)-f(x)}{2-x}\right\rvert=\left\lvert \frac{-f(x)}{2-x}\right\rvert>1$$ since $$f(x)>2-x$$. If there exists $$y\in (1,2)$$ such that $$f(y)<2-y$$ then again by the mean value theorem there exists $$v\in (1,y)$$ such that $$|f'(v)|=\left\lvert\frac{f(y)-f(1)}{y-1}\right\rvert=\left\lvert\frac{f(y)-1}{y-1}\right\rvert>1$$ since $$f(y)<2-y$$ so $$|f(y)-1|>|y-1|$$. So, on $$(1,2)$$ if the proposition does not hold then $$f(x)=2-x$$. However, notice that since $$f(x)$$ is differentiable at $$x=1$$ we cannot have $$f(x)=x$$ on $$(0,1)$$ and $$f(x)=2-x$$ on $$(1,2)$$. ◻

## (May 2016, 5)

Let $$f_n(x) = n^\beta x(1-x^2)^n$$, $$x \in [0,1]$$, $$n \in \mathbb{N}$$.

• Prove that $$\{f_n\}_{n=1}^\infty$$ converges pointwise on $$[0,1]$$ for every $$\beta \in \mathbb{R}$$.

• Show that the convergence in part (a) is uniform for all $$\beta < \frac{1}{2}$$, but not uniform for any $$\beta \geq \frac{1}{2}$$.

## (May 2016, 6)

• Suppose $$f \colon [-1,1] \to \mathbb{R}$$ is a bounded function that is continuous at $$0$$. Let $$\alpha(x) = -1$$ for $$x \in [-1,0]$$ and $$\alpha(x)=1$$ for $$x \in (0,1]$$. Prove that $$f \in \mathcal{R}(\alpha)[-1,1]$$, i.e., $$f$$ is Riemann integrable with respect to $$\alpha$$ on $$[-1,1]$$, and $$\int_{-1}^1 f d\alpha = 2f(0)$$.

• Let $$g \colon [0,1] \to \mathbb{R}$$ be a continuous function such that $$\int_0^1 g(x)x^{3k+2} dx = 0$$ for all $$k = 0, 1, 2, \ldots$$. Prove that $$g(x) =0$$ for all $$x \in [0,1]$$.

Proof. Let $$\epsilon>0$$. Choose $$\delta >0$$ so that if $$|x|<\delta$$, then $$|f(x)-f(0)|<\epsilon$$. Let $$P$$ be a partition of $$[-1,1]$$ with $$0 \in P$$ and $$\operatorname{mesh}(P)<\delta$$. Then $$|U(f,P,\alpha)-L(f,P,\alpha)|=|\sum_{i=1}^n(M_i-m_i)\Delta \alpha_i|=(|\sup_{x \in [0,x_k]}f(x)-\inf_{x \in [0,x_k]}f(x)|)2<4\epsilon$$. Thus $$f$$ is integrable with respect to $$\alpha$$. Additionally, we have $$L(f,P,\alpha)\leq 2f(0)\leq U(f,P,\alpha)$$ for all partitions $$P$$ of the form described above, and so $$\int_{-1}^1 f d\alpha = 2f(0)$$. ◻

Proof. Since $$g(x)$$ is continuous, so is $$g(x^{1/3})$$. Thus by the Weierstrauss Approximation Theorem, we can find a sequence of polynomials $$(p_n(x))\to g(x^{1/3})$$ uniformly. Since this holds for all values $$x\in [0,1]$$, we have that $$(p_n(x^3))$$ converges to $$g(x)$$ uniformly. Then we have $$(x^2p_n(x^3))$$ converges to $$x^2g(x)$$ uniformly. Note that by assumption, $$\int_0^1 g(x)x^2p_n(x^3)dx=0$$, and so $$0 = \lim_{n \to \infty}\int_0^1 g(x)x^2p_n(x^3)dx=\int_0^1 \lim_{n \to \infty}g(x)x^2p_n(x^3)dx=\int_0^1x^2g^2(x)dx$$. Since $$x^2g^2(x)$$ is non-negative, and its integral is zero, we conclude that $$x^2g^2(x)=0$$ for all $$x$$. Therefore, we have $$g(x)=0$$. ◻

# Metric Spaces and Topology

## (May 2019, 1)

Let $$(M, d_M)$$, $$(N, d_N)$$ be metric spaces. Define $$d_{M \times N} \colon (M \times N) \times (M \times N) \to \mathbb{R}$$ by \begin{align*}d_{M \times N}((x_1, y_1), (x_2, y_2)) := d_M(x_1, x_2) + d_N(y_1, y_2).\end{align*}

• Prove that $$(M \times N, d_{M \times N})$$ is a metric space.

• Let $$S \subseteq M$$ and $$T \subseteq N$$ be compact sets in $$(M, d_M)$$ and $$(N, d_N)$$, respectively. Prove that $$S \times T$$ is a compact set in $$(M \times N, d_{M \times N})$$.

Proof. To prove that $$(M \times N, d_{M \times N})$$ is a metric space we must prove that $$d_{M\times N}$$ is a metric on $$M \times N$$.

• Positive Definite-

Let $$(x_1,y_1), (x_2,y_2) \in M \times N$$. Then since $$d_M$$ is a metric on $$M$$, then $$d_M(x_1,x_2)\geq 0$$ for all $$x_i,x_j \in M$$ and $$d_N$$ is a metric on $$N$$ and likewise $$d_N(y_1,y_2)\geq 0$$ for any $$y_i,y_j \in N.$$

Then by definition $$d_{M\times N}((x_1,y_1),(x_2,y_2))=d_M(x_1,x_2)+d_N(y_1,y_2)\geq 0 + 0 =0.$$ Hence since $$(x_1,y_1),(x_2,y_2)$$ are arbitrary, $$d_{M\times N}((x_1,y_1),(x_2,y_2))\geq 0$$ for all $$(x_i,y_i),(x_j,y_j)\in M \times N$$.

Suppose that $$d_{M \times N}((x_1,y_1),(x_2,y_2))=0.$$ By definition $$d_{M \times N}((x_1,y_1),(x_2,y_2))=d_M(x_1,x_2)+d_N(y_1,y_2)$$. Therefore $$d_M(x_1,x_2)+d_N(y_1,y_2)=0$$, since $$d_M, d_N$$ are metrics, then $$d_M(x_1,x_2)\geq 0, d_N(y_1,y_2)\geq 0$$, which implies that $$d_M(x_1,x_2)=d_N(y_1,y_2)=0$$ and also since they are metrics we have that $$x_1=x_2, y_1=y_2.$$ Hence, $$(x_1,y_1)=(x_2,y_2).$$

Now suppose that $$(x_1,y_1)=(x_2,y_2).$$ Then $$x_1=x_2, y_1=y_2$$ and for the metrics $$d_M, d_N$$ we would have $$d_M(x_1,x_2)=0, d_N(y_1,y_2)=0.$$ Thus $$d_{M \times N}((x_1,y_1),(x_2,y_2))=d_M(x_1,x_2)+d_N(y_1,y_2)=0+0=0$$.

Therefore $$d_{M \times N}((x_1,y_1),(x_2,y_2))=0$$ if and only if $$(x_1,y_1)=(x_2,y_2).$$

• Symmetric

Let $$(x_1,y_1), (x_2,y_2) \in M \times N$$. Then since $$d_M$$ is a metric on $$M$$, then $$d_M(x_1,x_2)=d_M(x_2,x_1)$$ for all $$x_i,x_j \in M$$ and $$d_N$$ is a metric on $$N$$ and likewise $$d_N(y_1,y_2)=d_N(y_2,y_1)$$ for any $$y_i,y_j \in N.$$ Therefore, \begin{align*}\begin{aligned} d_{M \times N}((x_1,y_1),(x_2,y_2))&=d_M(x_1,x_2)+d_N(y_1,y_2)\\ &=d_M(x_2,x_1)+d_N(y_2,y_1)\\ &=d_{M \times N}((x_2,y_2),(x_1,y_1)). \end{aligned}\end{align*}

• Triangle Inequality

Since $$d_M, d_N$$ are metrics then for all $$x_1,x_2,x_3 \in M, y_1,y_2,y_3 \in N$$ we have that $$d_M(x_1,x_2)\leq d_M(x_1,x_3)+d_M(x_3,x_2)$$ and that $$d_N(y_1,y_2)\leq d_N(y_1,y_3)+d_N(y_3,y_2).$$ Therefore, \begin{align*}\begin{aligned} d_{M \times N}((x_1,y_1),(x_2,y_2))&=d_M(x_1,x_2)+d_N(y_1,y_2)\\ d_M(x_1,x_2)+d_N(y_1,y_2) &\leq d_M(x_1,x_3)+d_M(x_3,x_2)+d_N(y_1,y_3)+d_N(y_3,y_2)\\ d_M(x_1,x_3)+d_M(x_3,x_2)+d_N(y_1,y_3)+d_N(y_3,y_2) &=d_M(x_1,x_3)+d_N(y_1,y_3)+d_M(x_3,x_2)+d_N(y_3,y_2)\\ d_M(x_1,x_3)+d_N(y_1,y_3)+d_M(x_3,x_2)+d_N(y_3,y_2)&=d_M((x_1,y_1),(x_3,y_3))+d_M((x_3,y_3),(x_2,y_2)). \end{aligned}\end{align*}

Hence $$d_{M \times N}((x_1,y_1),(x_2,y_2))\leq d_M((x_1,y_1),(x_3,y_3))+d_M((x_3,y_3),(x_2,y_2)).$$

Therefore $$d_{M \times N}$$ is a metric on $$M \times N$$ and $$(M \times N, d_{M\times N})$$ is a metric space. ◻

Proof. By part a we showed that $$(M \times N, d_{M \times N})$$ is a metric space. Let $$\{s_n,t_n\}$$ be a sequence in $$S \times T.$$ Since $$\{s_n\}$$ is a sequence on a compact set $$S$$ in a metric space $$(M,d_M)$$ then it has a convergent subsequence $${s_{n_k}}.$$ Let $$\lim_{k \to \infty}{s_{n_k}}=s_0.$$

Since $$\{t_{n_k}\}$$ is a sequence on a compact set $$T$$ in a metric space. Thus $$\{t_{n_k}\}$$ has a convergent subsequence $$\{t_{n_{k_j}}\}.$$ Let $$\lim_{j\to \infty} t_{n_{k_j}}=t_0.$$ Thus $$\{s_{n_{k_j}}\}$$ is a subsequence of $$\{s_{n_k}\}.$$ And since $$\{s_{n_k}\}$$ converges to $$s_0$$, then any subsequence also converges to $$s_0.$$

Let $$\epsilon >0$$ be given. Then for $$\epsilon/2$$ there exists $$N_1, N_2\in \mathbb{N}$$ such that for all $$n_{k_j}\geq N_1, d_M(s_{n_{k_j}},s_0)<\epsilon/2$$, and for all $$n_{k_j}\geq N_2, d_N(t_{n_{k_j}},t_0)<\epsilon/2$$. Choose $$N=\text{Max}(\{N_1,N_2\}).$$

Then $$d_{M \times N}((s_{n_{k_j}},t_{n_{k_j}}),(s_0,t_0))=d_M(s_{n_{k_j}},s_0)+d_N(t_{n_{k_j}},t_0)<\epsilon/2 + \epsilon/2 = \epsilon.$$

Therefore $$d_{M \times N}((s_{n_{k_j}},t_{n_{k_j}}),(s_0,t_0))< \epsilon.$$

Hence $$\{(s_{n_{k_j}},t_{n_{k_j}})$$ converges to $$(s_0,t_0).$$ Therefore $$S \times T$$ is sequentially compact and $$S \times T$$ is therefore compact. ◻

## (June 2003, 1b,c)

• Show by example that the union of infinitely many compact subsets of a metric space need not be compact. (c) If $$(X,d)$$ is a metric space and $$K\subset X$$ is compact, define $$d(x_0,K)=\inf_{y\in K} d(x_0,y)$$. Prove that there exists a point $$y_0\in K$$ such that $$d(x_0,K)=d(x_0,y_0)$$.

## (January 2009, 4a)

Consider the metric space $$(\mathbb{Q},d)$$ where $$\mathbb{Q}$$ denotes the rational numbers and $$d(x,y)=|x-y|$$. Let $$E=\{x\in\mathbb{Q}:x>0,\,2<x^2<3\}$$. Is $$E$$ closed and bounded in $$\mathbb{Q}$$? Is $$E$$ compact in $$\mathbb{Q}$$?

## (January 2011 3a)

Let $$(X,d)$$ be a metric space, $$K\subset X$$ be compact, and $$F\subset X$$ be closed. If $$K\cap F=\emptyset$$, prove that there exists an $$\epsilon>0$$ so that $$d(k,f)\geq \epsilon$$ for all $$k\in K$$ and $$f\in F$$.

Proof. We prove this by contrapositive. Suppose for all $$\epsilon >0$$, there exists $$k \in K$$, $$f \in F$$ such that $$d(k,f)< \epsilon$$. Then for all $$n \in \mathbb{N}$$, we can choose $$k_n \in K$$, $$f_n \in F$$ such that $$d(k_n, f_n) < \frac{1}{n}$$.

Since $$k_n$$ is a sequence in $$K$$, which is compact (and therefore sequentially compact), there exists a subsequence $$k_{n_j} \subseteq k_n$$ with the property that $$k_{n_j}$$ converges to some $$k_0 \in K$$. Find $$N \in \mathbb{N}$$ such that for $$n \geq N$$, $$d(k_{n_j}, k_0) < \frac{\epsilon}{2}$$ and $$\frac{1}{n} < \frac{\epsilon}{2}$$. Then \begin{align*}d(f_{n_j}, k_0) \leq d(f_{n_j}, k_{n_j}) + d(k_{n_j}, k_0) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\end{align*}

Thus, $$f_{n_j}$$ also converges to $$k_0$$, and since $$F$$ is closed, $$k_0 \in F$$. So $$K \cap F \neq \emptyset$$. ◻

## 5?

Let $$(X,d)$$ be an unbounded and connected metric space. Prove that for each $$x_0 \in X$$, the set $$\{x \in X \, \colon \, d(x,x_0) = r\}$$ is nonempty.

# Sequences and Series

## (June 2013 1a)

Let $$a_n =\sqrt{n}\left(\sqrt{n+1}-\sqrt{n}\right)$$. Prove that $$\lim_{n\to\infty}a_n=1/2$$.

## (January 2014 2)

• Produce sequences $$\{a_n\},\,\{b_n\}$$ of positive real numbers such that \begin{align*}\liminf_{n\to\infty}(a_nb_n)>\left(\liminf_{n\to\infty} a_n\right) \left(\liminf_{n\to\infty} b_n\right).\end{align*}

• If $$\{a_n\},\,\{b_n\}$$ are sequences of positive real numbers and $$\{a_n\}$$ converges, prove that \begin{align*}\liminf_{n\to\infty}(a_nb_n)=\left(\lim_{n\to\infty}a_n\right)\left(\liminf_{n\to\infty}b_n\right).\end{align*}

## (May 2011 4a)

Determine the values of $$x\in\mathbb{R}$$ for which $$\displaystyle\sum_{n=1}^\infty \frac{x^n}{1+n|x|^n}$$ converges, justifying your answer carefully.

## (June 2005 3b)

If the series $$\sum_{n=0}^\infty a_n$$ converges conditionally, show that the radius of convergence of the power series $$\sum_{n=0}^\infty a_nx^n$$ is 1.

## (January 2011 5)

Suppose $$\{a_n\}$$ is a sequence of positive real numbers such that $$\lim_{n\to\infty}a_n=0$$ and $$\sum a_n$$ diverges. Prove that for all $$x>0$$ there exist integers $$n(1)<n(2)<\ldots$$ such that $$\sum_{k=1}^\infty a_{n(k)}=x$$.\

(Note: Many variations on this problem are possible including more general rearrangements. You may also wish to show that if $$\sum a_n$$ converges conditionally then given any $$x\in\mathbb{R}$$ there is a rearrangement of $$\{b_n\}$$ of $$\{a_n\}$$ such that $$\sum b_n=r$$. See Rudin Thm. 3.54 for a further generalization.)

## (June 2008 # 4b)

Assume $$\beta >0$$, $$a_n>0$$, $$n=1,2,\ldots$$, and the series $$\sum a_n$$ is divergent. Show that $$\displaystyle \sum \frac{a_n}{\beta + a_n}$$ is also divergent.

# Differential Calculus

## (June 2005 1a)

Use the definition of the derivative to prove that if $$f$$ and $$g$$ are differentiable at $$x$$, then $$fg$$ is differentiable at $$x$$.

## (January 2006 2b)

Assume that $$f$$ is differentiable at $$a$$. Evaluate \begin{align*}\lim_{x\to a}\frac{a^nf(x)-x^nf(a)}{x-a},\quad n\in\mathbb{N}.\end{align*}

## (June 2007 3a)

Suppose that $$f,g:\mathbb{R}\to\mathbb{R}$$ are differentiable, that $$f(x)\leq g(x)$$ for all $$x\in\mathbb{R}$$, and that $$f(x_0)=g(x_0)$$ for some $$x_0$$. Prove that $$f'(x_0)=g'(x_0)$$.

## (June 2008 3a)

Prove that if $$f'$$ exists and is bounded on $$(a,b]$$, then $$\lim_{x\to a^+}f(x)$$ exists.

## (January 2012 4b, extended)

Let $$f:\mathbb{R}\to\mathbb{R}$$ be a differentiable function with $$f'\in C(\mathbb{R})$$. Assume that there are $$a,b\in\mathbb{R}$$ with $$\lim_{x\to\infty}f(x)=a$$ and $$\lim_{x\to\infty}f'(x)=b$$. Prove that $$b=0$$. Then, find a counterexample to show that the assumption $$\lim_{x\to\infty}f'(x)$$ exists is necessary.

## (June 2012 1a)

Suppose that $$f:\mathbb{R}\to\mathbb{R}$$ satisfies $$f(0)=0$$. Prove that $$f$$ is differentiable at $$x=0$$ if and only if there is a function $$g:\mathbb{R}\to\mathbb{R}$$ which is continuous at $$x=0$$ and satisfies $$f(x)=xg(x)$$ for all $$x\in\mathbb{R}$$.

# Integral Calculus

## (January 2006 4b)

Suppose that $$f$$ is continuous and $$f(x)\geq 0$$ on $$[0,1]$$. If $$f(0)>0$$, prove that $$\int_0^1 f(x)dx>0$$.

## (June 2005 1b)

Use the definition of the Riemann integral to prove that if $$f$$ is bounded on $$[a,b]$$ and is continuous everywhere except for finitely many points in $$(a,b)$$, then $$f\in\mathscr{R}$$ on $$[a,b]$$.

## (January 2010 5)

Suppose that $$f:[a,b]\to\mathbb{R}$$ is continuous, $$f\geq 0$$ on $$[a,b]$$, and put $$M=\sup\{f(x):x\in[a,b]\}$$. Prove that \begin{align*}\lim_{p\to\infty}\left(\int_a^b f(x)^p\,dx\right)^{1/p}=M.\end{align*}

## (January 2009 4b)

Let $$f$$ be a continuous real-valued function on $$[0,1]$$. Prove that there exists at least one point $$\xi\in[0,1]$$ such that $$\int_0^1 x^4 f(x)\,dx=\frac{1}{5}f(\xi)$$.

Proof. Assume that $$f$$ is a continuous real-valued function on $$[0,1]$$. Then, by the Intermediate Value Theorem we have that $$f$$ attains its maximum and minimum on $$[0,1]$$. That is, for some $$a,b\in[0,1]$$,

We now have $$f(a)\leq f(x)\leq f(b)$$ for all $$x\in[0,1]$$. This gives \begin{align*}f(a)\int_0^1 x^4dx\leq \int_0^1 x^4f(x)dx\leq f(b)\int_0^1 x^4dx.\end{align*}

By the Fundamental Theorem of Calculus we know that

\begin{align*}\int_0^1x^4dx=\frac{1}{5}.\end{align*}

Thus, it follows that

\begin{align*}\frac{1}{5}f(a)\leq\int_0^1 x^4f(x)dx\leq \frac{1}{5}f(b)\end{align*} giving

\begin{align*}f(a)\leq 5\int_0^1 x^4f(x)dx\leq f(b).\end{align*}

By the Intermediate Value Theorem, there exists $$\xi\in[0,1]$$ such that

\begin{align*}f(\xi)=5\int_0^1 x^4f(x)dx.\end{align*}

Therefore, we have that there exists $$\xi\in[0,1]$$ such that $$\int_0^1 x^4 f(x)dx=\frac{1}{5}f(\xi)$$. ◻

## (June 2009 5b)

Let $$\phi$$ be a real-valued function defined on $$[0,1]$$ such that $$\phi$$, $$\phi'$$, and $$\phi''$$ are continuous on $$[0,1]$$. Prove that \begin{align*}\int_0^1 \cos x \frac{x\phi'(x)-\phi(x)+\phi(0)}{x^2}\,dx<\frac{3}{2}||\phi''||_\infty,\end{align*} where $$||\phi''||_\infty = \sup_{[0,1]}|\phi''(x)|.$$ Note that $$3/2$$ may not be the smallest possible constant.

# Sequences and Series of Functions

## (June 2010 6a)

Let $$f:[0,1]\to\mathbb{R}$$ be continuous with $$f(0)\neq f(1)$$ and define $$f_n(x)=f(x^n)$$. Prove that $$f_n$$ does not converge uniformly on $$[0,1]$$.

## (January 2008 5a)

Let $$f_n(x) = \frac{x}{1+nx^2}$$ for $$n \in \mathbb{N}$$. Let $$\mathcal{F} := \{f_n \, \colon \, n = 1, 2, 3, \ldots\}$$ and $$[a,b]$$ be any compact subset of $$\mathbb{R}$$. Is $$\mathcal{F}$$ equicontinuous? Justify your answer.

## (January 2005 4, June 2010 6b)

If $$f:[0,1]\to\mathbb{R}$$ is continuous, prove that \begin{align*}\displaystyle\lim_{n\to\infty}\int_0^1 f(x^n)\,dx=f(0).\end{align*}

## (January 2020 4a)

Let $$M<\infty$$ and $$\mathcal{F} \subseteq C[a,b]$$. Assume that each $$f \in \mathcal{F}$$ is differentiable on $$(a,b)$$ and satisfies $$|f(a)| \leq M$$ and $$|f'(x)| \leq M$$ for all $$x \in (a,b)$$. Prove that $$\mathcal{F}$$ is equicontinuous on $$[a,b]$$.

## (June 2005 5)

Suppose that $$f\in C([0,1])$$ and that $$\displaystyle \int_0^1 f(x)x^n\,dx=0$$ for all $$n=99,100,101,\ldots$$. Show that $$f\equiv 0$$.\

Note: Many variations on this problem exist. See June 2012 6b and others.

## (January 2005 3b)

Suppose $$f_n:[0,1]\to\mathbb{R}$$ are continuous functions converging uniformly to $$f:[0,1]\to\mathbb{R}$$. Either prove that $$\displaystyle\lim_{n\to\infty}\int_{1/n}^1 f_n(x)\,dx=\int_0^1 f(x)\,dx$$ or give a counterexample.

# Miscellaneous Topics

## (January 2018)

Let $$f \colon [a,b] \to \mathbb{R}$$. Suppose $$f \in \text{BV}[a,b]$$. Prove $$f$$ is the difference of two increasing functions.

## (January 2007, 6a)

Let $$f$$ be a function of bounded variation on $$[a,b]$$. Furthermore, assume that for some $$c>0$$, $$|f(x)| \geq c$$ on $$[a,b]$$. Show that $$g(x) = 1/f(x)$$ is of bounded variation on $$[a,b]$$.

## (January 2017, 2a)

Define $$f \colon [0,1] \to [-1,1]$$ by \begin{align*}f(x):= \begin{cases} x\sin\big({\frac{1}{x}}\big) & 0 < x \leq 1 \\ 0 & x = 0 \end{cases}\end{align*} Determine, with justification, whether $$f$$ is if bounded variation on the interval $$[0,1]$$.

## (January 2020, 6a)

Let $$\{a_n\}_{n=1}^\infty \subseteq \mathbb{R}$$ and a strictly increasing sequence $$\{x_n\}_{n=1}^\infty \subseteq (0,1)$$ be given. Assume that $$\sum_{n=1}^\infty a_n$$ is absolutely convergent, and define $$\alpha \colon [0,1] \to \mathbb{R}$$ by \begin{align*}\alpha(x):= \begin{cases} a_n & x=x_n \\ 0 & \text{otherwise} \end{cases}.\end{align*} Prove or disprove: $$\alpha$$ has bounded variation on $$[0,1]$$.

## Metric Spaces and Topology

• Find an example of a metric space $$X$$ and a subset $$E \subseteq X$$ such that $$E$$ is closed and bounded but not compact.

## (May 2017 6)

Let $$(X,d)$$ be a metric space. A function $$f \colon X \to \mathbb{R}$$ is said to be lower semi-continuous (l.s.c) if $$f^{-1}(a,\infty) = \{x \in X \, \colon \, f(x)> a\}$$ is open in $$X$$ for every $$a \in \mathbb{R}$$. Analogously, $$f$$ is upper semi-continuous (u.s.c) if $$f^{-1}(-\infty, b) = \{x \in X \, \colon \, f(x)<b\}$$ is open in $$X$$ for every $$b \in \mathbb{R}$$.

• Prove that a function $$f \colon X \to \mathbb{R}$$ is continuous if and only if $$f$$ is both l.s.c. and u.s.c.

• Prove that $$f$$ is lower semi-continuous if and only if $$\liminf_{n \to \infty} f(x_n) \geq f(x)$$ whenever $$\{x_n\}_{n=1}^\infty \subseteq X$$ such that $$x_n \to x$$ in $$X$$.

## (January 2017 3)

Let $$(X,d)$$ be a compact metric space. Suppose that $$f_n \colon X \to [0,\infty)$$ is a sequence of continuous functions with $$f_n(x) \geq f_{n+1}(x)$$ for all $$n \in \mathbb{N}$$ and $$x \in X$$, and such that $$f_n \to 0$$ pointwise on $$X$$. Prove that $$\{f_n\}_{n=1}^\infty$$ converges uniformly on $$X$$.

# Integral Calculus

## 1.

(June 2014 1)Define $$\alpha \colon [-1,1] \to \mathbb{R}$$ by \begin{align*}\alpha(x) := \begin{cases} -1 & x \in [-1,0] \\ 1 & x \in (0,1]. \end{cases}\end{align*} Let $$f \colon [-1,1] \to \mathbb{R}$$ be a function that is uniformly bounded on $$[-1,1]$$ and continuous at $$x=0$$, but not necessarily continuous for $$x \neq 0$$. Prove that $$f$$ is Riemann-Stieltjes integrable with respect to $$\alpha$$ over $$[-1,1]$$ and that \begin{align*}\int_{-1}^1 f(x)d\alpha(x) = 2f(0).\end{align*}

## (June 2017 2)

Prove : $$f \in \mathcal{R}(\alpha)$$ on $$[a,b]$$ if and only if for any $$a <c<b$$, $$f \in \mathcal{R}(\alpha)$$ on $$[a,c]$$ and on $$[c,b]$$. In addition, if either condition holds, then we have that \begin{align*}\int_a^c fd\alpha + \int_c^b fd\alpha = \int_a^b fd\alpha.\end{align*}

## (Spring 2017 7)

Prove that if $$f \in \mathcal{R}$$ on $$[a,b]$$ and $$\alpha \in C^1[a,b]$$, then the Riemann integral $$\int_a^b f(x)\alpha'(x)dx$$ exists and \begin{align*}\int_a^b f(x) d\alpha(x)= \int_a^b f(x)\alpha'(x)dx.\end{align*}

# Sequences and Series (and of Functions)

## (January 2006 1)

Let the power series series $$\sum_{n=0}^\infty a_nx^n$$ and $$\sum_{n=0}^\infty b_nx^n$$ have radii of convergence $$R_1$$ and $$R_2$$, respectively.

## ?

If $$R_1 \neq R_2$$, prove that the radius of convergence, $$R$$, of the power series $$\sum_{n=0}^\infty (a_n+b_n)x^n$$ is $$\min\{R_1, R_2\}$$. What can be said about $$R$$ when $$R_1 = R_2$$?

## ?

Prove that the radius of convergence, $$R$$, of $$\sum_{n=0}^\infty a_nb_nx^n$$ satisfies $$R \geq R_1R_2$$. Show by means of example that this inequality can be strict.

## ?

Show that the infinite series $$\sum_{n=0}^\infty x^n2^{-nx}$$ converges uniformly on $$[0,B]$$ for any $$B > 0$$. Does this series converge uniformly on $$[0,\infty)$$?

## (January 2006 4a)

Let \begin{align*}f_n(x) = \begin{cases} \frac{1}{n} & x \in (\frac{1}{2^{n+1}}, \frac{1}{2^n}] \\ 0 & \text{ otherwise}.\end{cases}\end{align*}

Show that $$\sum_{n=1}^\infty f_n$$ does not satisfy the Weierstrass M-test but that it nevertheless converges uniformly on $$\mathbb{R}$$.

## ?

Let $$f_n \colon [0,1) \to \mathbb{R}$$ be the function defined by \begin{align*}f_n(x):= \sum_{k=1}^n \frac{x^k}{1+x^k}.\end{align*}

• Prove that $$f_n$$ converges to a function $$f \colon [0,1) \to \mathbb{R}$$.

• Prove that for every $$0 < a < 1$$ the convergence is uniform on $$[0,a]$$.

• Prove that $$f$$ is differentiable on $$(0,1)$$.

## January 2019 Qualifying Exam

• Suppose that $$f: [0,1] \to \mathbb{R}$$ is differentiable and $$f(0) = 0$$. Assume that there is a $$k > 0$$ such that \begin{align*}|f'(x)| \leq k|f(x)|\end{align*} for all $$x\in [0,1]$$. Prove that $$f(x) = 0$$ for all $$x\in [0,1]$$.

Proof. Let $$0<\delta_1<1$$, and fix $$x_1 \in (0, \delta_1]$$. Since $$f(x)$$ is differentiable on all of $$[0,1]$$, $$f(x)$$ is differentiable on all of $$(0, \delta_1)$$. So by the Mean Value Theorem, there exists $$x_2 \in (0, x_1)$$ such that \begin{align*}f'(x_2) = \frac{f(x_1) - f(0)}{x_1-0} = \frac{f(x_1)}{x_1} .\end{align*} Solving for $$f(x_1)$$, we get $$f(x_1) = f'(x_2)x_1$$. So by hypothesis, $$f(x_1) = f'(x_2) x_1 \leq k|f(x_2)|x_1$$. Assume for $$x_1, x_2, \ldots, x_{n-1} \in (0,1)$$ the following conditions are satisfied for $$j\in\{1,2,\ldots, n-1\}$$. \begin{align*}\begin{aligned} x_j &\in& (0,x_{j-1}) \\ f(x_{j-1}) &=& f'(x_j)x_{j-1} \\ f(x_1) &\leq& k^{j-1}|f(x_j)|(x_{j-1} \cdots x_2x_1) \end{aligned}\end{align*} I now claim that this inductive process is true for $$j=n$$, given that it holds for all $$j \leq n$$. We apply the Mean Value Theorem to find some $$x_n \in (0, x_{n-1})$$ such that $$f'(x_n) = \frac{f(x_{n-1})}{x_{n-1}}$$, then write $$f(x_{n-1}) = f'(x_n)x_{n-1}.$$ By our inductive hypothesis, we have \begin{align*}\begin{aligned} |f(x_1)| &\leq& k^{n-2}|f(x_{n-1})|(x_{n-2}\cdots x_2x_1) \\ &=& k^{n-2}|f'(x_n)x_{n-1}|(x_{n-2}\cdots x_2x_1) \\ &\leq& k^{n-2}(k|f(x_n)|)(x_{n-1}x_{n-2}\cdots x_2x_1) \\ &=& k^{n-1}|f(x_n)|(x_{n-1}x_{n-2}\cdots x_2x_1). \end{aligned}\end{align*}

Thus our claim holds by induction. Now, since $$f$$ is a continuous function on the closed interval, we can apply the Extreme Value Theorem to find some $$M>0$$ for which $$f(x) \leq M$$ for all $$x\in [0,1]$$. Thus we get \begin{align*}|f(x_1)| \leq k^n M (x_n \cdots x_1) =(kx_n)(kx_{n-1})\cdots(kx_1)M\end{align*} for all $$n \in \mathbb{N}$$. If $$k < \frac{1}{x_1}$$, then for any $$\epsilon > 0$$ we can find $$N\in \mathbb{N}$$ sufficiently large so that $$|f(x_1)| < \epsilon$$. Otherwise, we set $$\delta_1< \frac{1}{k}$$ so that $$kx_1< 1$$. ◻