Prove that $|cdot|_1$ and $|cdot|_2$ are not equivalent norms on $C[0, 1]$.
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Consider the sequence $f_n(x) = x^n$. Then $|f_n|_1 = frac{1}{n+1}$, and $|f_n|_2= frac{1}{sqrt{2n+1}}$
How do you get from $f_n(x) = x^n$ to $|f_n|_1 = frac{1}{n+1}$ ?
functional-analysis
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add a comment |
$begingroup$
Consider the sequence $f_n(x) = x^n$. Then $|f_n|_1 = frac{1}{n+1}$, and $|f_n|_2= frac{1}{sqrt{2n+1}}$
How do you get from $f_n(x) = x^n$ to $|f_n|_1 = frac{1}{n+1}$ ?
functional-analysis
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2
$begingroup$
Welcome to MSE! What is the definition of $Vert cdot Vert_1$? Of $Vert cdot Vert_2$?
$endgroup$
– mathcounterexamples.net
Jan 16 at 16:39
add a comment |
$begingroup$
Consider the sequence $f_n(x) = x^n$. Then $|f_n|_1 = frac{1}{n+1}$, and $|f_n|_2= frac{1}{sqrt{2n+1}}$
How do you get from $f_n(x) = x^n$ to $|f_n|_1 = frac{1}{n+1}$ ?
functional-analysis
$endgroup$
Consider the sequence $f_n(x) = x^n$. Then $|f_n|_1 = frac{1}{n+1}$, and $|f_n|_2= frac{1}{sqrt{2n+1}}$
How do you get from $f_n(x) = x^n$ to $|f_n|_1 = frac{1}{n+1}$ ?
functional-analysis
functional-analysis
edited Jan 16 at 17:16
Domo Jens
asked Jan 16 at 16:37
Domo JensDomo Jens
12
12
2
$begingroup$
Welcome to MSE! What is the definition of $Vert cdot Vert_1$? Of $Vert cdot Vert_2$?
$endgroup$
– mathcounterexamples.net
Jan 16 at 16:39
add a comment |
2
$begingroup$
Welcome to MSE! What is the definition of $Vert cdot Vert_1$? Of $Vert cdot Vert_2$?
$endgroup$
– mathcounterexamples.net
Jan 16 at 16:39
2
2
$begingroup$
Welcome to MSE! What is the definition of $Vert cdot Vert_1$? Of $Vert cdot Vert_2$?
$endgroup$
– mathcounterexamples.net
Jan 16 at 16:39
$begingroup$
Welcome to MSE! What is the definition of $Vert cdot Vert_1$? Of $Vert cdot Vert_2$?
$endgroup$
– mathcounterexamples.net
Jan 16 at 16:39
add a comment |
1 Answer
1
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votes
$begingroup$
In your case,
$$|f|_1=int_0^1 |f(t)|dt,$$
and
$$|f|_2=sqrt{int_0^1 |f(t)|^2dt}.$$
$endgroup$
$begingroup$
I know but in this case ,how integral x^n get into 1/n+1
$endgroup$
– Domo Jens
Jan 16 at 17:17
$begingroup$
The primitive function of $x^n$ is $frac{x^{n+1}}{n+1}$ !
$endgroup$
– S. Maths
Jan 16 at 17:22
$begingroup$
thank you so much!
$endgroup$
– Domo Jens
Jan 16 at 19:18
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
In your case,
$$|f|_1=int_0^1 |f(t)|dt,$$
and
$$|f|_2=sqrt{int_0^1 |f(t)|^2dt}.$$
$endgroup$
$begingroup$
I know but in this case ,how integral x^n get into 1/n+1
$endgroup$
– Domo Jens
Jan 16 at 17:17
$begingroup$
The primitive function of $x^n$ is $frac{x^{n+1}}{n+1}$ !
$endgroup$
– S. Maths
Jan 16 at 17:22
$begingroup$
thank you so much!
$endgroup$
– Domo Jens
Jan 16 at 19:18
add a comment |
$begingroup$
In your case,
$$|f|_1=int_0^1 |f(t)|dt,$$
and
$$|f|_2=sqrt{int_0^1 |f(t)|^2dt}.$$
$endgroup$
$begingroup$
I know but in this case ,how integral x^n get into 1/n+1
$endgroup$
– Domo Jens
Jan 16 at 17:17
$begingroup$
The primitive function of $x^n$ is $frac{x^{n+1}}{n+1}$ !
$endgroup$
– S. Maths
Jan 16 at 17:22
$begingroup$
thank you so much!
$endgroup$
– Domo Jens
Jan 16 at 19:18
add a comment |
$begingroup$
In your case,
$$|f|_1=int_0^1 |f(t)|dt,$$
and
$$|f|_2=sqrt{int_0^1 |f(t)|^2dt}.$$
$endgroup$
In your case,
$$|f|_1=int_0^1 |f(t)|dt,$$
and
$$|f|_2=sqrt{int_0^1 |f(t)|^2dt}.$$
answered Jan 16 at 16:42
S. MathsS. Maths
657116
657116
$begingroup$
I know but in this case ,how integral x^n get into 1/n+1
$endgroup$
– Domo Jens
Jan 16 at 17:17
$begingroup$
The primitive function of $x^n$ is $frac{x^{n+1}}{n+1}$ !
$endgroup$
– S. Maths
Jan 16 at 17:22
$begingroup$
thank you so much!
$endgroup$
– Domo Jens
Jan 16 at 19:18
add a comment |
$begingroup$
I know but in this case ,how integral x^n get into 1/n+1
$endgroup$
– Domo Jens
Jan 16 at 17:17
$begingroup$
The primitive function of $x^n$ is $frac{x^{n+1}}{n+1}$ !
$endgroup$
– S. Maths
Jan 16 at 17:22
$begingroup$
thank you so much!
$endgroup$
– Domo Jens
Jan 16 at 19:18
$begingroup$
I know but in this case ,how integral x^n get into 1/n+1
$endgroup$
– Domo Jens
Jan 16 at 17:17
$begingroup$
I know but in this case ,how integral x^n get into 1/n+1
$endgroup$
– Domo Jens
Jan 16 at 17:17
$begingroup$
The primitive function of $x^n$ is $frac{x^{n+1}}{n+1}$ !
$endgroup$
– S. Maths
Jan 16 at 17:22
$begingroup$
The primitive function of $x^n$ is $frac{x^{n+1}}{n+1}$ !
$endgroup$
– S. Maths
Jan 16 at 17:22
$begingroup$
thank you so much!
$endgroup$
– Domo Jens
Jan 16 at 19:18
$begingroup$
thank you so much!
$endgroup$
– Domo Jens
Jan 16 at 19:18
add a comment |
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$begingroup$
Welcome to MSE! What is the definition of $Vert cdot Vert_1$? Of $Vert cdot Vert_2$?
$endgroup$
– mathcounterexamples.net
Jan 16 at 16:39