$C[0,1], 0<p<1$ is not a norm linear space. What about $p = infty$ (i.e. supremum norm)? [closed]
$begingroup$
To Prove - $C[0,1], 0<p<1$ (with usual p-norm) is not a norm linear space.
My thought process - I know that the triangle inequality is violated. But I am not able to produce a counterexample because every time I try to construct an example, I violate that it must be continuous on $[0,1]$.
One more subtle point, most other similar proofs that I've done usually are with the $f = g$(or $x = y$ for the two points of Triangle Inequality). But here, taking $f = g$ will not provide a counterexample since for such a case, Triangle Inequality holds with equality.
To prove/disprove - $C[0,1], $p = inifinity$ $ (i.e. the norm defined the supremum of the absolute value of the function on the closed interval $[0,1]$.) is not a norm linear space.
My thought process - The proof for p>=1 uses Holder Inequality, Minkowski Inequality. I wish to know if that thing can simply be used for p = infinity also. Then, that's done.
real-analysis calculus functional-analysis norm normed-spaces
asked Jan 13 at 18:30
Chahat ChawlaChahat Chawla
162
$endgroup$
closed as unclear what you're asking by Nate Eldredge, Lord Shark the Unknown, Xander Henderson, Eevee Trainer, max_zorn Jan 14 at 7:13
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
To Prove - $C[0,1], 0<p<1$ (with usual p-norm) is not a norm linear space.
My thought process - I know that the triangle inequality is violated. But I am not able to produce a counterexample because every time I try to construct an example, I violate that it must be continuous on $[0,1]$.
One more subtle point, most other similar proofs that I've done usually are with the $f = g$(or $x = y$ for the two points of Triangle Inequality). But here, taking $f = g$ will not provide a counterexample since for such a case, Triangle Inequality holds with equality.
To prove/disprove - $C[0,1], $p = inifinity$ $ (i.e. the norm defined the supremum of the absolute value of the function on the closed interval $[0,1]$.) is not a norm linear space.
My thought process - The proof for p>=1 uses Holder Inequality, Minkowski Inequality. I wish to know if that thing can simply be used for p = infinity also. Then, that's done.
real-analysis calculus functional-analysis norm normed-spaces
asked Jan 13 at 18:30
Chahat ChawlaChahat Chawla
162
$endgroup$
closed as unclear what you're asking by Nate Eldredge, Lord Shark the Unknown, Xander Henderson, Eevee Trainer, max_zorn Jan 14 at 7:13
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
I don't understand the question. Can you write down exactly the norms you are talking about, and the precise statement(s) you are trying to (dis)prove?
$endgroup$
– Nate Eldredge
Jan 14 at 3:52
$begingroup$
@NateEldredge, p norm is the integration over the interval [0,1] , of the p the power of the absolute value of the function, this whole integral made to the power 1/p. Also, the p = infinity norm is the supremum of the absolute value of the function on the interval [0,1].
$endgroup$
– Chahat Chawla
Jan 14 at 6:22
$begingroup$
@ChahatChawla Take whatever counterexample you've come up with, and tweak it very slightly, and you'll have a continuous function.
$endgroup$
– user3482749
Jan 14 at 17:08
add a comment |
$begingroup$
To Prove - $C[0,1], 0<p<1$ (with usual p-norm) is not a norm linear space.
My thought process - I know that the triangle inequality is violated. But I am not able to produce a counterexample because every time I try to construct an example, I violate that it must be continuous on $[0,1]$.
One more subtle point, most other similar proofs that I've done usually are with the $f = g$(or $x = y$ for the two points of Triangle Inequality). But here, taking $f = g$ will not provide a counterexample since for such a case, Triangle Inequality holds with equality.
To prove/disprove - $C[0,1], $p = inifinity$ $ (i.e. the norm defined the supremum of the absolute value of the function on the closed interval $[0,1]$.) is not a norm linear space.
My thought process - The proof for p>=1 uses Holder Inequality, Minkowski Inequality. I wish to know if that thing can simply be used for p = infinity also. Then, that's done.
real-analysis calculus functional-analysis norm normed-spaces
asked Jan 13 at 18:30
Chahat ChawlaChahat Chawla
162
$endgroup$
To Prove - $C[0,1], 0<p<1$ (with usual p-norm) is not a norm linear space.
My thought process - I know that the triangle inequality is violated. But I am not able to produce a counterexample because every time I try to construct an example, I violate that it must be continuous on $[0,1]$.
One more subtle point, most other similar proofs that I've done usually are with the $f = g$(or $x = y$ for the two points of Triangle Inequality). But here, taking $f = g$ will not provide a counterexample since for such a case, Triangle Inequality holds with equality.
To prove/disprove - $C[0,1], $p = inifinity$ $ (i.e. the norm defined the supremum of the absolute value of the function on the closed interval $[0,1]$.) is not a norm linear space.
My thought process - The proof for p>=1 uses Holder Inequality, Minkowski Inequality. I wish to know if that thing can simply be used for p = infinity also. Then, that's done.
real-analysis calculus functional-analysis norm normed-spaces
real-analysis calculus functional-analysis norm normed-spaces
asked Jan 13 at 18:30
Chahat ChawlaChahat Chawla
162
asked Jan 13 at 18:30
Chahat ChawlaChahat Chawla
162
edited Jan 14 at 6:19
Chahat Chawla
asked Jan 13 at 18:30
Chahat ChawlaChahat Chawla
162
asked Jan 13 at 18:30
Chahat ChawlaChahat Chawla
162
asked Jan 13 at 18:30
Chahat ChawlaChahat Chawla
162
162
closed as unclear what you're asking by Nate Eldredge, Lord Shark the Unknown, Xander Henderson, Eevee Trainer, max_zorn Jan 14 at 7:13
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Nate Eldredge, Lord Shark the Unknown, Xander Henderson, Eevee Trainer, max_zorn Jan 14 at 7:13
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
I don't understand the question. Can you write down exactly the norms you are talking about, and the precise statement(s) you are trying to (dis)prove?
$endgroup$
– Nate Eldredge
Jan 14 at 3:52
$begingroup$
@NateEldredge, p norm is the integration over the interval [0,1] , of the p the power of the absolute value of the function, this whole integral made to the power 1/p. Also, the p = infinity norm is the supremum of the absolute value of the function on the interval [0,1].
$endgroup$
– Chahat Chawla
Jan 14 at 6:22
$begingroup$
@ChahatChawla Take whatever counterexample you've come up with, and tweak it very slightly, and you'll have a continuous function.
$endgroup$
– user3482749
Jan 14 at 17:08
add a comment |
1
$begingroup$
I don't understand the question. Can you write down exactly the norms you are talking about, and the precise statement(s) you are trying to (dis)prove?
$endgroup$
– Nate Eldredge
Jan 14 at 3:52
$begingroup$
@NateEldredge, p norm is the integration over the interval [0,1] , of the p the power of the absolute value of the function, this whole integral made to the power 1/p. Also, the p = infinity norm is the supremum of the absolute value of the function on the interval [0,1].
$endgroup$
– Chahat Chawla
Jan 14 at 6:22
$begingroup$
@ChahatChawla Take whatever counterexample you've come up with, and tweak it very slightly, and you'll have a continuous function.
$endgroup$
– user3482749
Jan 14 at 17:08
1
1
$begingroup$
I don't understand the question. Can you write down exactly the norms you are talking about, and the precise statement(s) you are trying to (dis)prove?
$endgroup$
– Nate Eldredge
Jan 14 at 3:52
$begingroup$
I don't understand the question. Can you write down exactly the norms you are talking about, and the precise statement(s) you are trying to (dis)prove?
$endgroup$
– Nate Eldredge
Jan 14 at 3:52
$begingroup$
@NateEldredge, p norm is the integration over the interval [0,1] , of the p the power of the absolute value of the function, this whole integral made to the power 1/p. Also, the p = infinity norm is the supremum of the absolute value of the function on the interval [0,1].
$endgroup$
– Chahat Chawla
Jan 14 at 6:22
$begingroup$
@NateEldredge, p norm is the integration over the interval [0,1] , of the p the power of the absolute value of the function, this whole integral made to the power 1/p. Also, the p = infinity norm is the supremum of the absolute value of the function on the interval [0,1].
$endgroup$
– Chahat Chawla
Jan 14 at 6:22
$begingroup$
@ChahatChawla Take whatever counterexample you've come up with, and tweak it very slightly, and you'll have a continuous function.
$endgroup$
– user3482749
Jan 14 at 17:08
$begingroup$
@ChahatChawla Take whatever counterexample you've come up with, and tweak it very slightly, and you'll have a continuous function.
$endgroup$
– user3482749
Jan 14 at 17:08
add a comment |
0
active
oldest
votes
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
I don't understand the question. Can you write down exactly the norms you are talking about, and the precise statement(s) you are trying to (dis)prove?
$endgroup$
– Nate Eldredge
Jan 14 at 3:52
$begingroup$
@NateEldredge, p norm is the integration over the interval [0,1] , of the p the power of the absolute value of the function, this whole integral made to the power 1/p. Also, the p = infinity norm is the supremum of the absolute value of the function on the interval [0,1].
$endgroup$
– Chahat Chawla
Jan 14 at 6:22
$begingroup$
@ChahatChawla Take whatever counterexample you've come up with, and tweak it very slightly, and you'll have a continuous function.
$endgroup$
– user3482749
Jan 14 at 17:08