I'm trying to prove next propisition:

Let $X$ be a Banach space in either of the norms $||\cdot||_{1}$ or $||\cdot||_{2}.$ Suppose that $||\cdot||_{1}\leq C||\cdot||_{2}$ for some $C>0.$ Prove that there is a $D$ with $||\cdot||_{2}\leq D||\cdot||_{1}.$

I was trying to prove that $X$ is Banach space respect both norm to use inverse mapping theorem (a consequence of Open Map Theorem) applied to identity map and get the desire inequality, until I realized that $X$ could be complete respect $||\cdot||_{2}$ but not necessarily with respect to the other.

I'm stuck proving this. Any kind of help is thanked in advanced.

  • $\begingroup$ The problem states that $X$ is a Banach space with both norms, so you can apply the open mapping theorem. $\endgroup$ – Aweygan Dec 11 '17 at 9:04
  • $\begingroup$ What I understand is that $X$ is Banach space, for example, with respect $||\cdot||_{1},$ but not necessarily with respect the other. Isn't it? .I think is necessary to check each case separate @Aweygan. $\endgroup$ – Squird37 Dec 11 '17 at 9:48
  • $\begingroup$ I see what you mean. But the way you are understanding the statement makes it false. Consider the space $\ell^1$ with it's usual norm for $\|\cdot\|_2$ and the $2$-norm for $\|\cdot\|_1$. Then the identity map is bounded (with norm $1$), but it is also compact, hence not (boundedly) invertible. $\endgroup$ – Aweygan Dec 11 '17 at 17:26
  • $\begingroup$ I've been thinking in your counterexample, but I don't get it clear. The identity map is bounded because of the inequality and is bijective, so It has a continuous inverse. Now, if $X=\ell^1$ then the identity is not compact, because $\ell^1$ has infinite dimension. How do you conclude that the inverse (identity) is not bounded? $\endgroup$ – Squird37 Dec 12 '17 at 23:23
  • $\begingroup$ So I decided to summarize my thoughts in an answer that I hope explains my thoughts more clearly. Please take a look and let me know if you have any more questions. $\endgroup$ – Aweygan Dec 15 '17 at 18:48

So there are two (or three) ways of viewing this problem: Assuming that $X$ is complete in both the $\|\cdot\|_1$ and $\|\cdot\|_2$ norms, or with one of the norms. But only one of these is correct.

If we consider the first case, that is where $X$ is complete with both $\|\cdot\|_1$ and $\|\cdot\|_2$, and $\|\cdot\|_1\leq C\|\cdot\|_2$ for some $C>0$, then the two norms are equivalent. Indeed, this means the identity map $I:(X,\|\cdot\|_2)\to(X,\|\cdot\|_1)$ is bounded and bijective, hence the result follows from the open mapping theorem.

The second case is incorrect. Indeed, consider the case $X=\ell^1$, with $\|x\|_1=\sum_k|x(k)|$ (with which $X$ is Banach) and $\|x\|_2=\left(\sum_k|x(k)|^2\right)^{1/2}$ (with which $X$ is not Banach). Then we have $\|\cdot\|_2\leq\|\cdot\|_1$, but if we consider the sequence $(x_n)$ in $\ell_1$ defined by $$x_n(k)=\sum_{j=1}^n\frac{1}{j}e_j(k),$$ (where $e_j(k)=\delta_{jk}$ are the coordinate functions) then we have $$\|x_n\|_2=\left(\sum_{k=1}^n\frac{1}{k^2}\right)^{1/2}\leq\frac{\pi}{\sqrt{6}}$$ for all $n$, while $$\|x_n\|_1=\sum_{k=1}^n\frac{1}{k}\to\infty$$ ass $n\to\infty$. This means precisely that there is no $D>0$ such that $\|\cdot\|_1\leq D\|\cdot\|_2$.

  • $\begingroup$ Amazing @Aweygan! Thanks so much for such answer! $\endgroup$ – Squird37 Dec 19 '17 at 8:08
  • $\begingroup$ You're welcome, glad to help! $\endgroup$ – Aweygan Dec 19 '17 at 14:36
  • $\begingroup$ Hi may I ask why is the identity map bounded in the first case? $\endgroup$ – Korn Feb 4 at 5:12
  • $\begingroup$ @catsadnfish We're considering the identity map on $X$ as a map of normed spaces $I_{21}:(X,\|\cdot\|_2)\to(X,\|\cdot\|_1)$. Then the inequality $\|\cdot\|_1\leq C\|\cdot\|_2$ is just a different way of stating that the map $I_{21}$ is bounded. $\endgroup$ – Aweygan Feb 4 at 8:27

So what you have by the. Condition is that the identity map is continuous. Now since with respect to both norm, the space is Banach , the identity map is open so, the inverse identity map, which is again the identity map from $(X,||.||_{2})$ to $(X,||.||_{1})$ is continous. Now the result follows because continuity and an operator being bounded is the same thing. This is sometimes known as two-norm lemma.

  • $\begingroup$ Thanks to answer @Riju. When $X$ is Banach with both norms is enough to follow your argument, but what happend when $X$ is Banach only in one of them? $\endgroup$ – Squird37 Dec 11 '17 at 9:52
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    $\begingroup$ See if both the norms are equivalent then this means that the identity map is a linear homeomorphism and completenes is preserved under linear homeomorphism, so any way if one is banach then the condition to hold in your question, the other has to be Banach. I hope you understand what I want to say $\endgroup$ – Riju Dec 11 '17 at 10:43
  • $\begingroup$ I think I get your point, but I would like then to find a counterexample when $X$ is Banach only in one norm. Any suggestions? $\endgroup$ – Squird37 Dec 12 '17 at 23:34

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