Is the product of L2 norm and L-infinity norm convex?
0
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Let $x in mathbb{R}^n$ be an vector, and $f(x) = ||x||_2 cdot ||x||_infty$, I was wondering if $f(x)$ is a convex function?
Thanks!
optimization norm
asked Dec 28 '18 at 8:39
user3138073user3138073
1758
$endgroup$
add a comment |
0
$begingroup$
Let $x in mathbb{R}^n$ be an vector, and $f(x) = ||x||_2 cdot ||x||_infty$, I was wondering if $f(x)$ is a convex function?
Thanks!
optimization norm
asked Dec 28 '18 at 8:39
user3138073user3138073
1758
$endgroup$
$begingroup$
Could you give some context? Why is this question interesting? Have you tried anything? Also it is not clear what $x$ is.
$endgroup$
– Michh
Dec 28 '18 at 21:04
$begingroup$
- No, it isn't.
$endgroup$
– A.Γ.
Dec 28 '18 at 21:24
$begingroup$
@Michh $x$ is a vector, I have edited my question.
$endgroup$
– user3138073
Dec 28 '18 at 22:20
$begingroup$
@A.Γ.thanks for the answer, but could you tell me why?
$endgroup$
– user3138073
Dec 28 '18 at 22:22
$begingroup$
One can find a counterexample: $x,yinBbb{R}^n$ (with some large enough $n$) and $lambdain[0,1]$ such that $f(lambda x+(1-lambda)y)>lambda f(x)+(1-lambda)f(y)$.
$endgroup$
– A.Γ.
Dec 29 '18 at 10:07
add a comment |
0
0
0
$begingroup$
Let $x in mathbb{R}^n$ be an vector, and $f(x) = ||x||_2 cdot ||x||_infty$, I was wondering if $f(x)$ is a convex function?
Thanks!
optimization norm
asked Dec 28 '18 at 8:39
user3138073user3138073
1758
$endgroup$
Let $x in mathbb{R}^n$ be an vector, and $f(x) = ||x||_2 cdot ||x||_infty$, I was wondering if $f(x)$ is a convex function?
Thanks!
optimization norm
optimization norm
asked Dec 28 '18 at 8:39
user3138073user3138073
1758
asked Dec 28 '18 at 8:39
user3138073user3138073
1758
edited Dec 28 '18 at 22:20
user3138073
asked Dec 28 '18 at 8:39
user3138073user3138073
1758
asked Dec 28 '18 at 8:39
user3138073user3138073
1758
asked Dec 28 '18 at 8:39
user3138073user3138073
1758
1758
$begingroup$
Could you give some context? Why is this question interesting? Have you tried anything? Also it is not clear what $x$ is.
$endgroup$
– Michh
Dec 28 '18 at 21:04
$begingroup$
- No, it isn't.
$endgroup$
– A.Γ.
Dec 28 '18 at 21:24
$begingroup$
@Michh $x$ is a vector, I have edited my question.
$endgroup$
– user3138073
Dec 28 '18 at 22:20
$begingroup$
@A.Γ.thanks for the answer, but could you tell me why?
$endgroup$
– user3138073
Dec 28 '18 at 22:22
$begingroup$
One can find a counterexample: $x,yinBbb{R}^n$ (with some large enough $n$) and $lambdain[0,1]$ such that $f(lambda x+(1-lambda)y)>lambda f(x)+(1-lambda)f(y)$.
$endgroup$
– A.Γ.
Dec 29 '18 at 10:07
add a comment |
$begingroup$
Could you give some context? Why is this question interesting? Have you tried anything? Also it is not clear what $x$ is.
$endgroup$
– Michh
Dec 28 '18 at 21:04
$begingroup$
- No, it isn't.
$endgroup$
– A.Γ.
Dec 28 '18 at 21:24
$begingroup$
@Michh $x$ is a vector, I have edited my question.
$endgroup$
– user3138073
Dec 28 '18 at 22:20
$begingroup$
@A.Γ.thanks for the answer, but could you tell me why?
$endgroup$
– user3138073
Dec 28 '18 at 22:22
$begingroup$
One can find a counterexample: $x,yinBbb{R}^n$ (with some large enough $n$) and $lambdain[0,1]$ such that $f(lambda x+(1-lambda)y)>lambda f(x)+(1-lambda)f(y)$.
$endgroup$
– A.Γ.
Dec 29 '18 at 10:07
$begingroup$
Could you give some context? Why is this question interesting? Have you tried anything? Also it is not clear what $x$ is.
$endgroup$
– Michh
Dec 28 '18 at 21:04
$begingroup$
Could you give some context? Why is this question interesting? Have you tried anything? Also it is not clear what $x$ is.
$endgroup$
– Michh
Dec 28 '18 at 21:04
$begingroup$
- No, it isn't.
$endgroup$
– A.Γ.
Dec 28 '18 at 21:24
$begingroup$
- No, it isn't.
$endgroup$
– A.Γ.
Dec 28 '18 at 21:24
$begingroup$
@Michh $x$ is a vector, I have edited my question.
$endgroup$
– user3138073
Dec 28 '18 at 22:20
$begingroup$
@Michh $x$ is a vector, I have edited my question.
$endgroup$
– user3138073
Dec 28 '18 at 22:20
$begingroup$
@A.Γ.thanks for the answer, but could you tell me why?
$endgroup$
– user3138073
Dec 28 '18 at 22:22
$begingroup$
@A.Γ.thanks for the answer, but could you tell me why?
$endgroup$
– user3138073
Dec 28 '18 at 22:22
$begingroup$
One can find a counterexample: $x,yinBbb{R}^n$ (with some large enough $n$) and $lambdain[0,1]$ such that $f(lambda x+(1-lambda)y)>lambda f(x)+(1-lambda)f(y)$.
$endgroup$
– A.Γ.
Dec 29 '18 at 10:07
$begingroup$
One can find a counterexample: $x,yinBbb{R}^n$ (with some large enough $n$) and $lambdain[0,1]$ such that $f(lambda x+(1-lambda)y)>lambda f(x)+(1-lambda)f(y)$.
$endgroup$
– A.Γ.
Dec 29 '18 at 10:07
add a comment |
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$begingroup$
Could you give some context? Why is this question interesting? Have you tried anything? Also it is not clear what $x$ is.
$endgroup$
– Michh
Dec 28 '18 at 21:04
$begingroup$
- No, it isn't.
$endgroup$
– A.Γ.
Dec 28 '18 at 21:24
$begingroup$
@Michh $x$ is a vector, I have edited my question.
$endgroup$
– user3138073
Dec 28 '18 at 22:20
$begingroup$
@A.Γ.thanks for the answer, but could you tell me why?
$endgroup$
– user3138073
Dec 28 '18 at 22:22
$begingroup$
One can find a counterexample: $x,yinBbb{R}^n$ (with some large enough $n$) and $lambdain[0,1]$ such that $f(lambda x+(1-lambda)y)>lambda f(x)+(1-lambda)f(y)$.
$endgroup$
– A.Γ.
Dec 29 '18 at 10:07