Are the Weak and Strong Norms Correct for These Function Distances?

  • Context: MHB 
  • Thread starter Thread starter evinda
  • Start date Start date
  • Tags Tags
    Functions
Click For Summary

Discussion Overview

The discussion revolves around calculating the distances between two functions, \(y_1(x)=0\) and \(y_2(x)=\frac{1}{100} \sin(1000x)\), in the context of weak and strong norms within the space \(C^1[0,1]\). Participants explore the definitions and calculations related to these norms.

Discussion Character

  • Technical explanation, Debate/contested

Main Points Raised

  • One participant presents calculations for both weak and strong norms, arriving at specific values for each norm.
  • Another participant questions the definitions of the weak and strong norms, suggesting that the weak norm appears to be stronger than the strong norm based on their understanding.
  • A third participant references their textbook to clarify the definitions of weak and strong norms, asserting that the weak norm is defined as the maximum of the function and its derivative, while the strong norm is simply the maximum of the function.
  • There is a suggestion that the confusion might stem from a possible typo in the definitions provided by the first participant.

Areas of Agreement / Disagreement

Participants express uncertainty regarding the definitions of the weak and strong norms, indicating a lack of consensus on whether the calculations are correct based on potentially conflicting definitions.

Contextual Notes

There are unresolved questions about the definitions of the weak and strong norms, as well as the implications of the calculations presented. The discussion reflects differing interpretations of these norms.

evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
Hello! (Wave)

In $C^1[0,1]$ calculate the distances between the functions $y_1(x)=0$ and $y_2(x)=\frac{1}{100} \sin(1000x)$ in respect to the weak and strong norm.That's what I have tried:Weak norm:

$||y_1-y_2||_w=\max_{0 \leq x \leq 1} |y_1(x)-y_2(x)|+ \max_{0 \leq x \leq 1} |y_1'(x)-y_2'(x) | \\=\max_{0 \leq x \leq 1} \left |0-\frac{1}{100} \sin(1000x) \right|+ \max_{0 \leq x \leq 1} \left |\frac{1000}{100} \cos(1000x) \right|= \frac{1}{100}+10=\frac{1001}{100}$

Strong norm:

$||y_1-y_2||_M= \max_{0 \leq x \leq 1} |y_1(x)-y_2(x)|=\max_{0 \leq x \leq 1} \left |y_1(x)-y_2(x) \right|= \\=\max_{0 \leq x \leq 1} \left |0-\frac{1}{100} \sin(1000x) \right|= \frac{1}{100} \max_{0 \leq x \leq 1} |\sin(1000x)|= \frac{1}{100}$

Could you tell me if that what I have tried is right? (Thinking)
 
Physics news on Phys.org
evinda said:
Hello! (Wave)

In $C^1[0,1]$ calculate the distances between the functions $y_1(x)=0$ and $y_2(x)=\frac{1}{100} \sin(1000x)$ in respect to the weak and strong norm.That's what I have tried:Weak norm:

$||y_1-y_2||_w=\max_{0 \leq x \leq 1} |y_1(x)-y_2(x)|+ \max_{0 \leq x \leq 1} |y_1'(x)-y_2'(x) | \\=\max_{0 \leq x \leq 1} \left |0-\frac{1}{100} \sin(1000x) \right|+ \max_{0 \leq x \leq 1} \left |\frac{1000}{100} \cos(1000x) \right|= \frac{1}{100}+10=\frac{1001}{100}$

Strong norm:

$||y_1-y_2||_M= \max_{0 \leq x \leq 1} |y_1(x)-y_2(x)|=\max_{0 \leq x \leq 1} \left |y_1(x)-y_2(x) \right|= \\=\max_{0 \leq x \leq 1} \left |0-\frac{1}{100} \sin(1000x) \right|= \frac{1}{100} \max_{0 \leq x \leq 1} |\sin(1000x)|= \frac{1}{100}$

Could you tell me if that what I have tried is right? (Thinking)
If your definitions of the weak and strong norms are correct then so are those answers. But the definitions look odd to me because the "weak" norm is stronger than the "strong" norm. :confused:
 
Opalg said:
If your definitions of the weak and strong norms are correct then so are those answers. But the definitions look odd to me because the "weak" norm is stronger than the "strong" norm. :confused:

According to my textbook:In the linear space $C^1[a,b]$, the norm that is defined as

$$||y||_w=\max_{a \leq x \leq b} |y(x)|+ \max_{a \leq x \leq b} |y'(x)|$$

is called weak norm.

The maximum norm that we define like that: $||y||_M=\max_{a \leq x \leq b} |y(x)| dx$ is called strong norm.
 
Opalg said:
If your definitions of the weak and strong norms are correct then so are those answers. But the definitions look odd to me because the "weak" norm is stronger than the "strong" norm. :confused:

So do you think that it is a typo? (Thinking)
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 34 ·
2
Replies
34
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
2
Views
1K