Equivalence of two norms; Definition and Theorem from Kress

Question

The following is from Linear Integral Equations by Kress.

1. Should definition $1.4$ read that "...if each sequence converging ${\color{red}{\text{to } x \in X}}$ with respect to the first norm also converges ${\color{red}{\text{to the same }x}}$ with respect to the second norm and vice versa."
2. In the Proof of Theorem $1.5$ for the second part, how to prove that if there is no $C \in \mathbb{R}$, then there exists a sequence $(\phi_n)$ with $\Vert \phi_n\Vert_a = 1$ and $\Vert \phi_n \Vert_b \geq n^2$? Is the following the right way to think about?

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We will prove that if there doesn't exist $c, C\in \mathbb{R}$ such that $$c\Vert{\phi}\Vert_a \leq \Vert{\phi}\Vert_b \leq C\Vert{\phi}\Vert_a$$ for all $\phi \in X$, then the two norms $\Vert{\cdot}\Vert_a$ and $\Vert{\cdot}\Vert_b$ are not equivalent. Since there doesn't exist constants $c$ and $C$, this means there exists a sequence of $\phi_n \in X$, such that $\dfrac{\Vert{\phi_n}\Vert_b}{\Vert{\phi_n}\Vert_a} \geq n^2$. Consider the sequence $\psi_n = \dfrac1n \cdot \dfrac{\phi_n}{\Vert{\phi_n}\Vert_a}$. We then have $\Vert{\psi_n}\Vert_a = \dfrac1n$ and $\Vert{\psi_n}\Vert_b \geq n$. This means $\lim_{n \to \infty} \Vert{\psi_n}\Vert_a = 0$, whereas $\lim_{n \to \infty} \Vert{\psi_n}\Vert_b \geq n$. Hence, $\psi_n \to 0$ in $\Vert{\cdot}\Vert_a$, whereas $\psi_n$ doesn't converge to $0$ in $\Vert{\cdot}\Vert_b$.

Answer 1

1 The definition of equivalent I've seen taken is the existence of the $c$, $C$. i.e. two norms $\|\cdot\|_a$ and $\|\cdot\|_b$ are equivalent if there are real constants $c, C>0$ such that for all $\phi$ in $X$ $$c\|\phi\|_a\leq\|\phi\|_b\leq C\|\phi\|_a$$

The author shows that this is equivalent to his definition of equivalent, with the extra fact that the limits coincide (in the Theorem), so I'd assume this is deliberately omitted from the definition (Why? I'm not sure without reading the text, sorry)

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2 Yes your proof is correct. It's worth mentioning the trivial $X$ case in my opinion. Below is a slightly reformulated version of the same proof.

Suppose there is no $C>0$ such that for all $\phi\in X$ $$\|\phi\|_b\leq C\|\phi\|_a$$ Note that $X$ isn't trivial, since then any $C$ would satisfy $0\leq C\cdot 0$. In particular for each $n$ there is some $\psi_n$ such that $$\|\psi_n\|_b\not\leq n^2{\|\psi_n\|_a}.$$ Now defining $$\phi_n:=\frac{\psi_n}{\|\psi_n\|_a}$$ Then $$\|\phi_n\|_a=\frac{\|\psi_n\|_a}{\|\psi_n\|_a}=1$$ Using the absolute homogeneity property. And $$\|\phi_n\|_b=\frac{\|\psi_n\|_b}{\|\psi_n\|_a}\geq n^2.$$

The result of the theorem picks up in Kress's proof. :-)