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Advanced Calculus - Continuous Functions
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In summary, Bradyrsmith31's problem with the proofs is that they are not correct and he does not believe Euge's proof.
- #2
Euge
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Hi bradyrsmith31,
For part (a), take any $x \in [0,1] \cap \Bbb Q$ and let $x_n := \frac{x}{2^{1/2^n}}$. Then $x_n \in [0,1]\setminus \Bbb Q$ such that $x_n \to x$. Since $f(x_n) = 0$ for all $n$ and $f(x) > 0$, $f(x_n)$ does not converge to $f(x)$. Therefore, $f$ is discontinuous at $x$.
For part (b), let $\varepsilon > 0$ and $x_0 \in [0,1] \setminus \Bbb Q$. Since $\lim_{n \to \infty} a_n = 0$, there exists a positive integer $N$ such that for all $n \ge N$, $a_n < \varepsilon$. Partition $[0,1]$ into $N$ subintervals $J_1,\ldots, J_N$ of length $\frac{1}{N}$. Let $k$ be the index such that $x_0 \in J_k$. Set $\delta$ equal to the distance from $x_0$ to the nearer endpoint of $J_k$. If $|x - x_0| < \delta$, then $x$ lies in the interior of $J_k$. If $x$ is rational, this implies $d(x) \ge N$. Consequently, $$|f(x) - f(x_0)| = |f(x) - 0| = f(x) = a_{d(x)} < \varepsilon.$$ If $x$ is irrational, then $|f(x) - f(x_0)| = |0 - 0| = 0 < \varepsilon$. Since $\varepsilon$ and $x_0$ are arbitrary, $f$ is continuous at every point of $[0,1]\setminus \Bbb Q$.
For part (c), note that $f$ is bounded and continuous outside a null set.
For part (d), consider any partition $P : 0 = x_0 < x_1 < \cdots < x_n = 1$ of $[0,1]$. The lower Riemann sum $L(f,P)$ is zero because each interval $[x_{i-1}, x_i]$ contains an irrational point $q_i$, and $f(q_i) = 0 \le \inf_{t \in [x_{i-1},x_i]} f(t)$. Consequently, the lower Riemann integral of $f$ is zero.
The upper Riemann integral of $f$ is also zero. For let $\varepsilon > 0$. Let $N$ be a positive integer such that $a_n < \frac{\varepsilon}{2}$ for all $n \ge N$. Partition $[0,1]$ into the intervals $I_k :=[\frac{k-1}{N}, \frac{k}{N}]$, for $k = 1, 2,\ldots, N$. For each $k$, let $M_k := \sup_{t \in I_k} f(t)$. There exist $r_1,\ldots, r_n$ such that $t_k \in I_k$ such that $M_k < f(r_k) + \frac{\varepsilon}{2}$ for all $k$. Given $k$, if $r_k$ is rational, $f(r_k) = 0$; if $r_k$ is irrational, $d(r_k) \ge N$ and thus $f(r_k) = a_{d(r_k)} < \frac{\varepsilon}{2}$. Thus $M_k < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$ for each $k$. The upper Riemann sum $U(f,P) = \frac{1}{n} \sum_{k = 1}^n M_k $ is therefore less than $\varepsilon$. Since $\varepsilon$ is arbitrary, this shows that the upper Riemann sum of $f$ is zero.
For part (a), take any $x \in [0,1] \cap \Bbb Q$ and let $x_n := \frac{x}{2^{1/2^n}}$. Then $x_n \in [0,1]\setminus \Bbb Q$ such that $x_n \to x$. Since $f(x_n) = 0$ for all $n$ and $f(x) > 0$, $f(x_n)$ does not converge to $f(x)$. Therefore, $f$ is discontinuous at $x$.
For part (b), let $\varepsilon > 0$ and $x_0 \in [0,1] \setminus \Bbb Q$. Since $\lim_{n \to \infty} a_n = 0$, there exists a positive integer $N$ such that for all $n \ge N$, $a_n < \varepsilon$. Partition $[0,1]$ into $N$ subintervals $J_1,\ldots, J_N$ of length $\frac{1}{N}$. Let $k$ be the index such that $x_0 \in J_k$. Set $\delta$ equal to the distance from $x_0$ to the nearer endpoint of $J_k$. If $|x - x_0| < \delta$, then $x$ lies in the interior of $J_k$. If $x$ is rational, this implies $d(x) \ge N$. Consequently, $$|f(x) - f(x_0)| = |f(x) - 0| = f(x) = a_{d(x)} < \varepsilon.$$ If $x$ is irrational, then $|f(x) - f(x_0)| = |0 - 0| = 0 < \varepsilon$. Since $\varepsilon$ and $x_0$ are arbitrary, $f$ is continuous at every point of $[0,1]\setminus \Bbb Q$.
For part (c), note that $f$ is bounded and continuous outside a null set.
For part (d), consider any partition $P : 0 = x_0 < x_1 < \cdots < x_n = 1$ of $[0,1]$. The lower Riemann sum $L(f,P)$ is zero because each interval $[x_{i-1}, x_i]$ contains an irrational point $q_i$, and $f(q_i) = 0 \le \inf_{t \in [x_{i-1},x_i]} f(t)$. Consequently, the lower Riemann integral of $f$ is zero.
The upper Riemann integral of $f$ is also zero. For let $\varepsilon > 0$. Let $N$ be a positive integer such that $a_n < \frac{\varepsilon}{2}$ for all $n \ge N$. Partition $[0,1]$ into the intervals $I_k :=[\frac{k-1}{N}, \frac{k}{N}]$, for $k = 1, 2,\ldots, N$. For each $k$, let $M_k := \sup_{t \in I_k} f(t)$. There exist $r_1,\ldots, r_n$ such that $t_k \in I_k$ such that $M_k < f(r_k) + \frac{\varepsilon}{2}$ for all $k$. Given $k$, if $r_k$ is rational, $f(r_k) = 0$; if $r_k$ is irrational, $d(r_k) \ge N$ and thus $f(r_k) = a_{d(r_k)} < \frac{\varepsilon}{2}$. Thus $M_k < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$ for each $k$. The upper Riemann sum $U(f,P) = \frac{1}{n} \sum_{k = 1}^n M_k $ is therefore less than $\varepsilon$. Since $\varepsilon$ is arbitrary, this shows that the upper Riemann sum of $f$ is zero.
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- #3
johng1
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I don't believe Euge's proof is correct. He seems to be saying that if q is a rational between k/N and k/(N+1), d(q) is greater than or equal to N. Of course, this is false: 500/1001<1/2<501/1001. In particular, I don't believe his proof of b) in case $x_0$ is "very close" to 1/2.
Here is my proof of the problem. The idea of the proof of c) and d) is fairly simple, but the details are a little messy.
Here is my proof of the problem. The idea of the proof of c) and d) is fairly simple, but the details are a little messy.
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- #4
Euge
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Hi johng,
You're right about my proof of (b) (and part of (d)). Even in the proof of (d), I mixed up the words "rational" and "irrational" near the end. I meant to fix these issues later tonight, but you beat me to it! In any case, thanks for the critique.
You're right about my proof of (b) (and part of (d)). Even in the proof of (d), I mixed up the words "rational" and "irrational" near the end. I meant to fix these issues later tonight, but you beat me to it! In any case, thanks for the critique.
- #5
blue_raver22
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I understand how difficult and frustrating it can be to tackle advanced calculus proofs. However, with persistence and the right approach, these proofs can be solved.
Firstly, it is important to have a strong understanding of the definition of a continuous function. A function f(x) is continuous at a point a if the following three conditions are met:
1. f(a) is defined
2. The limit of f(x) as x approaches a exists
3. The limit of f(x) as x approaches a is equal to f(a)
Using this definition, you can begin to approach proofs involving continuous functions by breaking them down into smaller, more manageable steps.
One useful strategy is to start by assuming the function is continuous at a given point and then working backwards to prove that the three conditions are met. This approach allows you to focus on each condition individually and use the properties of continuous functions to make logical deductions.
Another important tool for solving continuous function proofs is the use of epsilon-delta arguments. These arguments involve choosing an arbitrary value for epsilon (ε) and then finding a corresponding value for delta (δ) that satisfies the definition of continuity. This approach can be useful for proving the existence of limits and showing that the limit is equal to f(a).
It is also important to have a strong understanding of the properties of continuous functions, such as the intermediate value theorem, the extreme value theorem, and the composition of continuous functions. These properties can often be used to simplify proofs and make them more manageable.
In conclusion, tackling advanced calculus proofs involving continuous functions requires a combination of understanding the definition of continuity, using logical deductions, and applying relevant properties. With practice and perseverance, you will be able to successfully solve these proofs.
FAQ: Advanced Calculus - Continuous Functions
What are continuous functions?
Continuous functions are mathematical functions that have no abrupt changes or breaks in their graph. This means that as the input values of the function change, the output values also change smoothly without any sudden jumps or gaps.
What is the importance of continuity in calculus?
Continuity is a crucial concept in calculus because it allows us to make predictions and calculations about the behavior of a function over a certain interval. Without continuity, many important theorems and formulas in calculus would not hold true.
How do you determine if a function is continuous?
A function is continuous if it satisfies the three conditions of continuity: 1) the function is defined at the point in question, 2) the limit of the function as the input approaches the point exists, and 3) the limit is equal to the function value at the point.
Can a function be continuous at one point but not at others?
Yes, a function can be continuous at one point but not at others. This is known as a point discontinuity, where the function is continuous everywhere except at a specific point. This is typically caused by a discontinuity or sharp turn in the graph at that point.
How are continuous functions used in real-world applications?
Continuous functions are widely used in various fields, such as physics, engineering, economics, and more. They are used to model and predict real-world phenomena, such as fluid flow, stock market trends, and population growth. The concept of continuity is also essential in optimization problems, where we seek to find the maximum or minimum value of a function.