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)=x1x2, x(0,1).

  • By using the ϵδ 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.

proof:

Fix x(0,1) and let ϵ>0. Then we have |f(x)f(y)|=|x1x2y1y2|=|x(1y2)y(1x2)(1x2)(1y2)|=|xy(1x)(1+x)(1y)(1+y)|. Now, choose δ>0 such that δ<min. 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:

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:

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:

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:

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:

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:

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:

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:

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.

Continuity of Functions

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:

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],

\begin{align*}f(a)=\min\limits_{[0,1]}f(x) \qquad \text{and} \qquad f(b)=\max\limits_{[0,1]}f(x).\end{align*}

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

Bounded Variation

(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:

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. ◻