$||.||_1\leq a||.||_2$. Do the cases $ a∈(0,1) $and $a∈[1,∞)$ matter to be open sets under these norms?

Question

If there are two norms $||.||_1$ and $ ||.||_2 $ on a vector space V such that $||.||_1\leq a||.||_2$ for some positive real number $a$, then can we say that each open set in $(V,||.||_2)$ is also open in $(V,||.||_1)$? But how? Does the value of $a$ matter? Do the cases $a\in(0,1) $ and $a\in[1,\infty)$ matter?

Answer 1

One way to show that one topology $\sigma$ is equal to another $\tau$ if you know that $\mathcal B $ is a base for the former and $\mathcal C$ is a base for the latter is to show that $\mathcal B = \mathcal C$.

In the case of normed spaces, the norm topology has a basis of balls of arbitrary radii. If $B_i(r,x)$ is the ball of radius $r$ about a point $x$ with respect to the ith norm then: $$B_1(r,x) = B_2(r/a,x)$$ $$B_2(r,x) = B_1(ar,x)$$

Hence ball in the first norm is a ball of some radius in the second and vice versa. There bases coincide, so the two topologies are equal (ie any open set in one topology is open in the other).

You should hopefully be able to see that the value of $a$ is irrelevant so long as it is positive.

Answer 2

If $V$ is a Banach space under both of your norms, then it holds as a Corollary of the Open Mapping Theorem and the value of $a>0$ doesn't matter.

If $V$ is not a Banach space, I guess you could find some simple counterexamples in $c_{00}$, but here are some more general examples that where easier for me to come up with. In what follows, every $V$ is infinite dimensional.

1. Let $(x_i)_{i\in I}$ be a normalized Hamel basis of the Banach space $(V, \|\cdot\|)$ and for every $x=\sum_{i=1}^n\lambda_i x_i$, define $\|x\|'=\sum_{i=1}^n|\lambda_i|$. It's easy to check that $\|\cdot\|'$ defines indeed a norm on $V$ and that for every $x\in V$, $$\|x\|=\big\|\sum_{i=1}^n\lambda_ix_i\big\|\leq \sum_{i=1}^n|\lambda_i|\|x_i\|=\sum_{i=1}^n|\lambda_i|=\|x\|'.$$ So the two norms satisfy the desired inequality. Notice also that $(V, \|\cdot\|')$ is not a Banach space. A quick way to see this is to check that under the $\|\cdot\|'$ norm, all the coordinate functionals $(x_i^\#)_{i\in I}$ are bounded, while in Banach spaces at most finitely many of them can be bounded. Now, since $V$ under its original norm is a Banach space, again from the same property, there exists a coordinate functional, say $x_i^\#$ which is not $\|\cdot\|$-bounded, so its kernel $\ker x_i^\#$ is not $\|\cdot\|$-closed. But, $\ker x_i^\#$ is $\|\cdot\|'$-closed since each $x_j^\#$ is $\|\cdot\|'$-bounded.

2. The second one is similar but much shorter and comes from here. Let $(V, \|\cdot\|)$ be a Banach space, pick a discontinuous linear $f\in V^*$ and define a norm $\|\cdot\|'$ as $\|x\|'=\|x\|+|f(x)|$, for $x\in V$. Then $\|x\|\leq \|x\|'$, for every $x\in V$ and it should be easy to check that $\ker f$ is $\|\cdot\|'$-closed but not $\|\cdot\|$-closed. Here the original space $(V, \|\cdot\|)$ doesn't even need to be a Banach space, all you need is an infinite dimensional space, so that you can find a discontinuous functional.

3. Consider $c_{00}$ equipped with the $\ell_1$ and $\ell_2$ norms. It's trivial to check that $\|x\|_2\leq \|x\|_1$, for every $x\in c_{00}$. Since $(c_{00}, \|\cdot\|_1)^*=\ell_\infty$ and $(c_{00}, \|\cdot\|_2)^*=\ell_2\subsetneq \ell_\infty$, there exists an $f$ which is $\|\cdot\|_1$ continuous but not $\|\cdot\|_2$ continuous. Its kernel gives a counterexample. For instance, the set $F=\{(x_n)\in c_{00}: \sum x_n=0 \}$ is $\|\cdot\|_1$-closed, but $\|\cdot\|_2$-dense.

There is a common trend that appeared in all the constructions above: The two norms gave different duals and to find a counterexample we just needed to pick a functional which was continuous only with respect to one of the norms. This is not coincidental:

Proposition: Let $\|\cdot\|_1$, $\|\cdot\|_2$ be norms on $X$ such that $(X, \|\cdot\|_1)^*$ coincides with $(X, \|\cdot\|_2)^*$. Then the two norms on $X$ are equivalent.

<p><strong>Proof:</strong> 
In any normed space $Y$, an application of the Uniform Boundedness Principle yields that a set $A\subseteq Y$ is bounded if and only if $y^*(A)$ is a bounded subset of $\mathbb{R}$ for every $y^*\in Y^*$. Since in our space the two duals are the same, this property implies that a set $A\subseteq X$ is $ \|\cdot\|_1$-bounded  if and only if it is $ \|\cdot\|_2$-bounded. </p>

<p>Suppose that the two norms are not equivalent. Then there exists a sequence $(x_n)_{n\in\mathbb{N}}$ in $X$ such that $\|x_n\|_1=1$ and $\|x_n\|_2\geq n$, for every $n\in \mathbb{N}$, a contradiction since $(x_n)_{n\in\mathbb{N}}$ is bounded only in the $\|\cdot\|_1$ norm.  Similarly you obtain a contradiction for the opposite inequality.</p>

So our only hope to find a counterexample, was to consider two norms which do not give the same dual. Actually any such space will provide us with one:

Proposition: Let $\|\cdot\|_1, \|\cdot\|_2$ be two norms on $X$ for which $\|x\|_1\leq C\|x\|_2$, for some constant $C>0$, for every $x\in X$. The following are equivalent:

<p>1) The spaces $(X, \|\cdot\|_1)$ and $(X, \|\cdot\|_2)$ have the same open sets.</p>

<p>2) The spaces $(X, \|\cdot\|_1)^*$ and $(X, \|\cdot\|_2)^*$ coincide.</p>

<p><strong>Proof:</strong> 
The direction $2) \Rightarrow 1)$ is the previous Proposition. For the converse, suppose that $(X, \|\cdot\|_1)$ and $(X, \|\cdot\|_2)$ have different duals. Then you can pick an $f$ which is $\|\cdot\|_2$-bounded but not $\|\cdot\|_1$-bounded and then its kernel shows that property 1) does not hold.</p>