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

[cbarker1]

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?

Relevant Equations

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.

 

[andrewkirk]

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.