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--- math104-s21:s:ryotainagaki:problems [2021/05/08 04:22]
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+++ math104-s21:s:ryotainagaki:problems [2026/02/21 14:41] (current)
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 $$|\frac{x + y}{x^2 + y^2} - \frac{2}{2}| = |\frac{x+y-x^2-y^2}{2(x^2+y^2)}| = |\frac{x(1- x) + y(1-y)}{x^2 + y^2}|$$
-We intend to prove the continuity through the $\epsilon - \delta$ definition. So we want to find $\delta > 0$ where $\forall d((x, y), (1, 1)) < \delta, |f(x, y) - f(1, 1)| < \epsilon$.
+We intend to prove the continuity through the $\epsilon - \delta$ definition. So we want to find $\delta > 0$ where $\forall d(<x , y>, <1 , 1>) < \delta, |f(x, y) - f(1, 1)| < \epsilon$.
-Using the $d((x, y), (1, 1)) < \delta$, we can say that $|y-1| < \delta$ and $|x-1| < \delta$.
+Using the $d(<x, y>, <1,  1>) < \delta$, we can say that $|y-1| < \delta$ and $|x-1| < \delta$.
 Using clever inequalities and upper bounds, we can say... (assuming that our delta is going to be smaller than one,
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 That is a positive number on the right hand side (you may want to prove that as a VERY CHALLENGING exercise).
-In other words, we know $\forall \epsilon > 0, \exists \delta > 0$, here it can be that $$\delta < \min(\frac{2}{1 + 2\sqrt{2}}, \frac{-(1 + 2\epsilon\sqrt{2}) + \sqrt{(4\sqrt{2} + 8)\epsilon + 2}}{2(1 - \epsilon)})$$,  |f(x, y) -f(1, 1)| < \epsilon$.
+In other words, we know $\forall \epsilon > 0, \exists \delta > 0: \forall <x, y> \in \mathbb{R^2}, (\sqrt{(x - 1)^2 + (y - 1)^2} < \delta) \to   (|f(x, y) -f(1, 1)| < \epsilon)$.
 Thus by the $\epsilon, \delta$ definition of continuity, we know that $f(x, y)$ is continuous at $(x, y)$.
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-** 32. Give an example of a function that is continuous on $\mathbb{Q}$ but not on $\mathbb{R}$ and RIGOROUSLY prove that your functions satisfies such property. **
+** 32. (Credits to Midterm 2) In less than 3 sentences and in less than 3 minutes, formally prove or disprove: given $A \subseteq B$ is bounded and f is a continuous map $f: B \to Y$, then $f(A)$ is bounded. **
-**Answer: ** Consider the function $f(x)$ defined as $1$ if $x \in \mathbb{Q}$ and $0$ otherwise. We can say that on $\mathbb{Q}$ the function is rational since on $\mathbb{Q}$ $f(x) = 1$. However on $\mathbb{R}$ the function is not continuous since for any point x_0 in $\mathbb{R}$, $\lim_{x \to x_0}f(x)$ does not exist. Why? Using a sequence $x_n$ of irrational numbers, I get that $f(x_n) = 0 $ and $ f(x_n) \to 0$. Using a sequence $x_n$ of rational numbers, I get that $f(x_n) = 1$ and $ f(x_n) \to 1$.
+**Answer: On the spot and on a timed fast exam, this may seem like a hard problem and deceptive. ** We know that the given statement is false. To show that consider the metric space $(S, d(x, y )=|x- y|)$, $U = (0, 1)$, and $f(x) = \ln(x)$. We know that $U$ is bounded but $f(U) = (-\infty, 0)$ by the property of natural log and is NOT bounded. Thus, just because $A$ is bounded doesn't mean that $f(A)$ is bounded.
+** 33. Suppose that (a_n)_n is a sequence in a metric space (M, d), which converges to a limit
+$a \in M$. Prove that $K = \{a_n | n \in \mathbb{N}\} \cup {a}$ is compact. **
-** 33. Suppose that (an)n is a sequence in a metric space (M, d), which converges to a limit
+** Answer: **
-$a \in M$. Prove that $K = \{an | n \in \mathbb{N}} \cup {a}$ is compact. **
+Firstly, consider an arbitrary open cover $\{C_{\alpha}\}$ of K. Since $\{C_{\alpha}\}$ is an open cover of $K$, we know that there exists an open set $S$ in the open cover that contains $a$. Since $S$ contains a and since $S$ is open, we know that $\exists r > 0: \{x \in M: d(x, a) < r\} \subseteq S$. Because $a_n \to a$, we know that $\exists N \mathbb{N}: \forall n > N, d(a_n, a) < r$. This means that $\{a_n: n > N\} \subseteq \{x \in M: d(x, a) < r\} \subseteq S$. Now consider the first $N$ points in $a_n$. In order for the open cover $\{C_{\alpha}\}$ to cover $\{a_n: n \in \{1, 2, ... , N\}\}$, one needs only at most $N$ sets from $\{C_{\alpha}\}$ whose union contain $a_1, ..., a_N$. Therefore, there exists a finite number of sets that cover the union of a) $\{a\}\cup \{a_n: n > N\}$ and b) finite set $\{a_n: n \in \{1, ... , N\}\}$. Thus, there exists a finite subcover of $\{C_{\alpha}\}$. By definition of of compact, $K$ is compact.
 ** 34. Problem 5 on Practice Midterm on [[https://ywfan-math.github.io/104s21_midterm2_practice.pdf]] **
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 ** 44. Does $f_N(x) = \sum_{i=0}^{N} \frac{1}{x^2n^2 + 1}\sqrt{\frac{1}{x}}$ converge uniformly on $(1, \infty)$? Support your claim with a proof. **
-** Answer: ** Consider that for $x \in (1, \infty), 0 \leq \frac{1}{x^2n^2 + 1}\sqrt{\frac{1}{x}} \leq \frac{1}{n^2 + 1} \leq \frac{1}{n^2}$. Since we know by the P-Series test that $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges, we know that $\sum_{n=0}^{\infty}\frac{1}{x^2n^2 + 1}\sqrt{\frac{1}{x}}$ converges by comparison test regardless of what $x \in (1, \infty)$ is.
+** Answer: ** Consider that $\forall x \in (1, \infty), |\frac{1}{x^2n^2 + 1}\sqrt{\frac{1}{x}}| \leq \frac{1}{n^2 + 1}\sqrt{1} = \frac{1}{n^2}$
-Since $\sum_{n = 0}^{\infty}\sum_{n=0}^{\infty}\frac{1}{x^2n^2 + 1}\sqrt{\frac{1}{x}}$ converges regardless of what $x \in (1, \infty)$ is we know that $F_N(x) = \frac{1}{x^2n^2 + 1}\sqrt{\frac{1}{x}}$ is cauchy for all $x \in (1, \infty)$.
+We know that $\sum_{n=1}^{\infty}\frac{1}{n^2}$ converges by the P series test. Thus, by the Weitrass-M test, we know that $f_N(x) = \sum_{i=0}^{N} \frac{1}{x^2n^2 + 1}\sqrt{\frac{1}{x}$ converges uniformly to some function $f$.
-Thus we know that $(x \in (1, \infty))$ implies $F_N$ is cauchy. By cauchy criterion for uniform convergence $f_N$ converges uniformly on $(1, \infty)$
 ** 45. (From Dr. Ian Charlesworth's Fall 2020 Midterm) Given $A, B \subset \mathbb{R}$ and let A be bounded from above and B be bounded from below. Let $A - B = \{s - t: s\ in A, t \in B\}$. Prove that $\sup{(A - B)} = \sup A - \inf B$. **
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-** 49. Complete the proof of L'Hopital's theorem in the April 8 notes (morning) by considering the case where A is infinite. **
+** 49. From Ian Charlesworth's Fall 2020 Final  (Illustrative question) Without invoking the change of variables, show that $\int_{x=C}^{x = 2C}f(\frac{x}{C})dx$ is integrable and find the integral. We are given that $\int_{x=1}^{x=2}f(x)dx$ is a valid integral. (let h(x) = f(x/C) for $x \in [C, 2C]$ and note that $C > 0$)**
+**Answer: ** This really tests our understanding of the definition of integrability.
+We know that since $\int_{x=1}^{x=2}f(x)dx = V$ means that $\forall \epsilon. > 0, \exists P: U(f, P) - L(f, P) < \epsilon $. Recall that $$U(f, P) = \sum_{i=1}^{n}\sup_{x \in [x_{i-1}, x_i]} f(x) (x_i - x_{i-1})$$ and $$L(f, P) = \sum_{i=1}^{n}\inf_{x \in [x_{i-1}, x_i]} f(x) (x_i - x_{i-1})$$. Now consider that for each partition$ P = \{1 = x_0 < x_1 < x_2 ... < x_n = 2\}$ of $[1, 2]$ we can create a partition of $[C, 2C]$ by doing $P* = \{C = Cx_0 < Cx_1 < ... < Cx_n = 2\}$. Now consider $$U(h, P*) = \sum_{i=1}^{n}\sup_{x \in [Cx_{i-1}, Cx_i]} h(x) (Cx_i - Cx_{i-1})= \sum_{i=1}^{n}\sup_{x \in [Cx_{i-1}, Cx_i]} f(x/C) (Cx_i - Cx_{i-1})$$ and $$L(h, P*) = \sum_{i=1}^{n}\sup_{x \in [Cx_{i-1}, Cx_i]} h(x) (Cx_i - Cx_{i-1}) = \sum_{i=1}^{n}\inf_{x \in [Cx_{i-1}, Cx_i]} f(x/C) (Cx_i - Cx_{i-1})$$.
+Notice that because of the x/C used in h(x), we can effectively say that $\sup_{x \in [Cx_{i-1}, Cx_{i}]}f(x/C) = \sup_{x \in [x_{i-1}, x_{i}]}f(x)$. Therefore,  $$U(h, P*) = \sum_{i=1}^{n}\sup_{x \in [x_{i-1}, x_i]} f(x) (Cx_i - Cx_{i-1}) = C U(h, P)$$ and $$L(h, P*)  = \sum_{i=1}^{n}\inf_{x \in [x_{i-1}, x_i]} f(x) (Cx_i - Cx_{i-1}) = C L(h, P)$$.
+Therefore, $$U(h, P*) - L(h, P*) < C(U(f, P) - L(f, P)) < C\epsilon$$.
+In order to make $$U(h, P*) - L(h, P*) < \epsilon$$ we can first find $P = \{1 = x_0 < x_1 < ... , < x_n = 2\}$ such that $U(f, P) - L(h, P) < \epsilon/C$ and then we can make $P* = \{C = x_0 < Cx_1 < ... < Cx_n = 2C}$ and that way $U(h, P*) - L(h, P*) = C (\cdot U(f, P) - \cdot L(f, P)) < C \frac{\epsilon}{C} = \epsilon $.
+Thus, $\forall \epsilon > 0, \exists P*$ (partition of $[C, 2C]$) such that $U(h, P*) - L(h, P*) < \epsilon$. Thus $h$ is integrable on $[C, 2C]$.
+We have shown that $U(h, P*) = C U(f, P)$, $L(h, P*) = C L(f, P)$. We know that $\inf U(f, P) = U(f)$. Since $U(h, P*) = C U(f, P)$ and how $P*$ is setup as $\{Cx_0 < C x_1 < .. < C x_n<\}$, I know $U(h) = \inf U(h, P*) = \inf C(U(f, P)) = C U(f)$. Likewise, we know that $\sup L(f, P) = L(f)$. Since $L(h, P*) = C L(f, P)$ and how $P*$ is setup as $\{Cx_0 < C x_1 < .. < C x_n<\}$, I know $L(h) = \inf L(h, P*) = \inf L(U(f, P)) = C L(f)$. Since $L(f) = U(f) = \int_{x=1}^{x=2}f(x)dx$ by definition of integrability, we know that $L(h) = U(h) = C \int_{x=1}^{x=2}f(x)dx$.
 ** 50. Give an example of a function that is Stieltjes Integrable but not Riemann Integrable. **
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 However, we cannot say that $f$ is Riemann integrable over $[-1, 1]$. This is due to the behaviour of $f$. And when we consider interval $[-1, 0]$ we start to understand that $f$ is not integrable because of the upper integral for that interval is 1 whereas the lower integral for that interval would be 0.
+** 51. (From MIT Opencourseweare Math 18.100B Course) True or False? If $f_n : [a, b] \to R$is a sequence of almost everywhere continuous functions, and $f_n \to f$ converges uniformly, then the limit f is almost everywhere continuous. (I wanted an explanation with the specific theorem numbers from Rudin)**
+**Answer: ** I kind of do not like how the term "almost everywhere continuous" is used in this problem; it just sounds imprecise. It turns out that the problem meant f has only a finite number of discontinuities.  In order to address this problem we may want to first note that since $f_n \to f$ continuous on $[a, b]$ except for a finite number of points on $[a, b]$ we know that $f_n$ is Riemann integrable ($\alpha(x) = x$ and that is continuous for all $x \in \mathbb{R}$) on $[a, b]$ by theorem 6.10. Because of that and by ** theorem 7.16 ** in Rudin, we can supposedly say that $f$ is integrable as well. And since f is integrable, f MUST have only finitely many discontinuities.(This one does not seem to have a direct theorem from Rudin I can use.)
+One thing that I am still rightfully confused about is whether it is true that if $f$ has only finite number of discontinuities then and only then it is riemann integrable. I may be able to puncture this argument by considering $f(x) = \frac{1}{x^2}$. We may all know that $f(x)$ is continuous at any point but at $x = 0$, where $\lim_{x \to 0} f(x) = \infty$. Now, we know that f has only a finite number (1) of discontinuities on $[-a, b]$ (let a, b be any positive number of your choosing), but we cannot integrate $\int_{x= -a}^{x = b}f(x)dx$. It will diverge to infinity (not a real number)!
+Sources for the Questions and Inspiration: Past midterms and/or finals from Dr. Peng Zhou, Ian Charlesworth, Sebastian Eterovic, Yu-Wei Fan, Dr. Charles Rycroft, MIT Opencourseware (MIT's Math 18.100 courses), summer 2018 math 104, spring 2017 math 104, //Principles of Analysis// by Rudin, //A Problem Book in Real Analysis//, and //Principles of Analysis// by Ross.

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