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

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The discussion revolves around calculating the weak and strong norms for the functions y1(x)=0 and y2(x)=1/100 sin(1000x) in the space C^1[0,1]. The weak norm is calculated as 1001/100, while the strong norm is determined to be 1/100. Participants express confusion over the definitions of the weak and strong norms, suggesting that the weak norm appears to be stronger than the strong norm, which contradicts standard definitions. A textbook definition is referenced, clarifying that the weak norm includes both function and derivative maxima, while the strong norm only considers the function maximum. The conversation concludes with a suggestion that there may be a typographical error in the definitions provided.
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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