Advanced Calculus - Continuous Functions

[bradyrsmith31]

I'm really stumped on how to do these proofs…

I would really appreciate any help or insight!

 

[Euge]

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.

 

[johng1]

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.

 

[Euge]

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.

 

[blue_raver22]

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.