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

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SUMMARY

The discussion centers on calculating the distances between the functions \(y_1(x)=0\) and \(y_2(x)=\frac{1}{100} \sin(1000x)\) in the space \(C^1[0,1]\) using weak and strong norms. The weak norm is defined as \(||y_1-y_2||_w=\max_{0 \leq x \leq 1} |y_1(x)-y_2(x)| + \max_{0 \leq x \leq 1} |y_1'(x)-y_2'(x)|\), yielding a result of \(\frac{1001}{100}\). The strong norm is defined as \(||y_1-y_2||_M=\max_{0 \leq x \leq 1} |y_1(x)-y_2(x)|\), resulting in \(\frac{1}{100}\). There is confusion regarding the definitions, as the weak norm appears to be stronger than the strong norm, which contradicts conventional terminology.

PREREQUISITES
  • Understanding of function spaces, specifically \(C^1[0,1]\)
  • Knowledge of weak and strong norms in functional analysis
  • Familiarity with calculus, particularly derivatives and maxima
  • Ability to interpret mathematical notation and definitions
NEXT STEPS
  • Review the definitions of weak and strong norms in functional analysis literature
  • Study the properties of \(C^1[a,b]\) spaces and their norms
  • Explore examples of calculating norms for different functions
  • Investigate potential typographical errors in mathematical definitions and their implications
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Mathematicians, students of functional analysis, and anyone interested in understanding norms in function spaces will benefit from this discussion.

evinda
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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)
 
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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)
 

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