General approach to proving that two norms are/aren’t equivalent

functional-analysisnormed-spacesvector-spaces

By definition two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent on a vector space $X$ iff there exist nonzero constants $\alpha$ and $\beta$ so that for all $x\in X$

$$\alpha||x||_1\leq||x||_2\leq\beta||x||_1$$

In layman's terms $||\cdot||_1$ and $||\cdot||_2$ are equivalent iff you can bound one of them by the other one and vice versa. If two norms are equivalent then it's usually not too difficult to find $\alpha$ and $\beta$ fitting the above mentioned inequalities. However I can't seem to understand how to find counterexamples in the form of vector $x$ for which one of the norms is equal to $0$ or $+\infty$ thus showing that at least one of the constants can't exist.

Let's consider the uniform norm $\|\cdot\|_{\infty}$ and the norm

$$||g||_{0}=\sup_{-\infty<y<+\infty}|g(y) \cdot \arctan(y)|$$ on the vector space of continuous functions on $\mathbb{R}$ with a compact support.

You can bound $||\cdot||_{0}$ with a constant $\beta=\frac{\pi}{2}$ from above but I don't believe you can do the same from below. How would you go about proving it?

Best Answer

You seem to be asking two things. I am replaying to your title and overall idea on how to prove when two norms are not equivalent.

Let $\mathrm{V}_i$ be the normed vector space $(\mathrm{X}, \| \cdot \|_i)$ for $i = 1, 2.$

Theorem. Let $I:\mathrm{X} \to \mathrm{X}$ denote the function $I(x) = x$ (that is, $I$ is the "identity"). The following conditions are equivalent:

  1. The identity $I:\mathrm{V}_1 \to \mathrm{V}_2$ is continuous at zero.

  2. For every ball $\mathrm{B}_2$ centred at zero in $\mathrm{V}_2,$ there exists a ball $\mathrm{B}_1$ centred at zero in $\mathrm{V}_1$ such that $I(\mathrm{B}_1) \subset \mathrm{B}_2.$

  3. The unit ball $\mathrm{B}_1$ of $\mathrm{V}_1$ has bounded image by $I.$

  4. The operator norm $\| I \| < \infty.$

  5. There is a constant $c > 0$ such that $\| x \|_2 \leq c \| x \|_1.$

Sketch of proof. We go by proving that item $i$ implies item $i+1$ (modulo 5). First, 2. is the definition of continuity at zero for the identity, so 1. implies 2. Next, if $2.$ holds, and $\mathrm{B}_i$ has radius $r_i,$ then $\dfrac{1}{r_1} \mathrm{B}_1$ is the unit ball of $\mathrm{V}_1$ and its image by $I$ is contained in the ball of radius $\dfrac{r_2}{r_1}$ in $\mathrm{V}_2.$ So, 2. implies 3. If 3. holds, let $c > 0$ be such that $\|I(x)\|_2 \leq c$ for all $x$ in the unit ball of $\mathrm{V}_1,$ then $\left\| I \left( \frac{x}{2\|x\|_1} \right) \right\|_2 \leq c,$ so $\|I(x)\|_2 \leq 2c \|x\|_1$ and $\| I \| < \infty$ by definition, so 4. holds. If 4. holds, then 5. holds immediately. If 5. holds, then as $x \to 0$ in $\mathrm{V}_1$ then $I(x) \to 0$ in $\mathrm{V}_2,$ and $I$ is continuous at zero. Q.E.D.

To your exercise. The previous theorem gives you analytical and geometrical interpretations of two norms being equivalent (the equivalence of the norms allows the previous theorem to apply in both directions). Break one of these and you prove the two norms are not equivalent. Often is easy to see that the unit ball of one norm cannot be rescaled to fit inside the unit ball of the other norm, hence breaking 3.

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