Does each norm on vector space become discontinuous when restricted to S^1?

In summary, the conversation discusses the proof of a problem involving two distinct norms on the same vector space over complex numbers. The conclusion is that there exists no real number K that satisfies a certain inequality. The conversation also touches on the continuity and convergence of the norms in question.
  • #1
cbarker1
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Homework Statement
Prove that if two norms on the same vector space are not equivalent, then at least one of them is discontinuous on the unit sphere in the other norm. Does each norm become discontinuous when restricted to the unit sphere of the other?
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Axioms of norm and Vector space
Dear Everybody,
I am having trouble with last part of this question.
I believe the answer is no. But I have to proof the general case. Here is my work for the problem:
Suppose that we have two distinct norms on the same vector space ##X## over complex numbers. Then there exists no ##K## in the real numbers so that ##\|\cdot\|_2\leq K\|\cdot\|_{1}.## Define a sequence of ##\|\cdot\|_{1}## unit vectors ##(x_n)_{n\in N}## so that ##\|x_n\|_2\geq 3^n.## Now ##\frac{x_1+\frac{1}{3nx_n}}{\|x_1+\frac{1}{3nx_n}\|_{1}}## is a sequence of ##\|\cdot\|_{1}## unit vectors converging to ##x_1## in ##\|\cdot \|_{1}.## But, the norm of the sequences is ##\bigg\|\frac{x_1+\frac{1}{3nx_{n}}}{\|x_1+\frac{1}{nx_{n}}\|_{1}}\bigg\|_{2}\geq \frac{|\frac{1}{3n}\|x_n\|_{2}-\|x_1\|_{2}|}{\|x_1+\frac{1}{3n}x_{n}\|_1}\geq \frac{|\frac{3^{n-1}}{n}-\|x_{1}\|_{2}|}{\|x_1+\frac{1}{3n}x_n\|_1}## does not converge in ##\|x_1\|_2,## so ##\|\cdot\|_2## is discontinuous on the unit sphere of ##\|\cdot\|_1.##
No. If we assume that ##\|\cdot\|_{1}\leq C\|\cdot\|_{2},## then we know that ##\|\cdot\|_{2}## Cauchy sequence converges in ##\|\cdot\|_{1}## and we know that ##\|\cdot\|_{1}## is continuous on the vector space ##X## and if we restricted ##\|\cdot\|_{1}## to the unit sphere.
 
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  • #2
cbarker1 said:
Define a sequence of ##\|\cdot\|_{1}## unit vectors ##(x_n)_{n\in N}## so that ##\|x_n\|_2\geq 3^n.##
We can't just define that, since we don't know whether any such sequence exists.
We can define a sequence of ##\|\cdot\|_1## unit vectors, or we can define a sequence of vectors whose ##\|\cdot\|_2## magnitude increases as ##3^n##. But we can't just assume the existence of a sequence that has both properties.
First we'd need to prove that such a sequence exists.
 
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Related to Does each norm on vector space become discontinuous when restricted to S^1?

What is a norm on a vector space?

A norm on a vector space is a function that assigns a non-negative length or size to each vector in the space, satisfying certain properties: positivity, scalar multiplication, the triangle inequality, and the property that the norm of the zero vector is zero.

What is S^1 in the context of vector spaces?

S^1 typically refers to the unit circle in a vector space, specifically in a two-dimensional real or complex vector space. It consists of all vectors with a norm (or length) of 1.

Why might a norm become discontinuous when restricted to S^1?

A norm might become discontinuous when restricted to S^1 if the norm function is not uniformly continuous on the entire vector space or if the topology induced by the norm on the vector space does not align well with the topology of the unit circle. This can happen in certain exotic or less common norms.

Are all norms continuous on the entire vector space?

Yes, by definition, all norms are continuous functions on their entire vector space. Discontinuity issues typically arise when restricting the norm to a subset like S^1 under specific conditions or in non-standard contexts.

Can you give an example of a norm that becomes discontinuous on S^1?

An example would be a norm defined in a non-standard or pathological manner, such as a norm that changes behavior drastically at certain points. However, in typical mathematical contexts involving standard norms (e.g., Euclidean, p-norms), norms remain continuous even when restricted to S^1.

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