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. Can each norm be discontinuous when restricted to the unit sphere of the other?

We know that if $\Vert\cdot\Vert$ and $\Vert\cdot\Vert'$ are not equivalent, then either for every $C>0$ there exists $x$ with $\Vert x\Vert>C\Vert x\Vert'$ or for every $C>0$ there exists $x$ with $\Vert x\Vert'>C\Vert x\Vert$.

Need some hints to proceed with the problem.


1 Answer 1


Suppose there exists no $C$ so that $\|\cdot\|_2≤C\,\|\cdot\|_1$. Then you can find a sequence of $\|\cdot\|_1$-unit vectors $x_n$ so that $\|x_n\|_2≥2^n$. Now $\frac{x_1+1/n x_n}{\|x_1 + 1/n x_n\|_1}$ is a sequence of $\|\cdot\|_1$-unit vectors converging to $x_1$ in $\|\cdot\|_1$. However the two norm of these guys is $$\left\|\frac{x_1+1/nx_n}{\|x_1+1/nx_n\|_1}\right\|_2≥\frac{\left|1/n\|x_n\|_2- \|x_1\|_2\right|}{\|x_1+1/n x_n\|_1}≥\frac{\left|2^n/n- \|x_1\|_2\right|}{\|x_1+1/n x_n\|_1}$$ which does not converge $\|x_1\|_2$, so $\|\cdot\|_2$ is discontinous on the unit sphere of $\|\cdot\|_1$.

If you have $\|\cdot\|_1≤ C\|\cdot\|_2$ however, then any $\|\cdot\|_2$ Cauchy Sequence is Cauchy on $\|\cdot\|_1$, and $\|\cdot\|_1$ is continuous on the entire vector space thus also on the unit sphere.

  • $\begingroup$ The unit sphere is not the same thing as the unit ball. $\endgroup$
    – PhoemueX
    Commented Aug 6, 2017 at 16:18
  • $\begingroup$ @PhoemueX oh, right. I rewrote it then. $\endgroup$
    – s.harp
    Commented Aug 7, 2017 at 8:21

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