Proof that two norms on Banach space are equivalent

Question

Let $V$ be a Banach space with respect to norms $\|\cdot\|_{1}$ and $\|\cdot\|_2$. There exists $Q>0$ such that $\forall_{v\in V} \|v\|_1\leq Q\|v\|_2$. Show that there exists $P>0$ such that $\forall_{v\in V} \|v\|_2\leq P\|v\|_1$.

I have tried proving it by contradiction, but then I would get a sequence $v_n$ such that $$ \forall_n \|v_n\|_2>n\|v_n\|_1$$ and I could not get anything from it. I guess that I need to find a Cauchy sequence that doesn't converge in order to get a contradiction. Because without assumption of completeness it's obviously false.

Answer 1

The identity map from $(V,\|.\|_2)$ to $ (V,\|.\|_1)$ is continuous and Open Mapping Theorem tells you that the identity map from $(V,\|.\|_1)$ to $ (V,\|.\|_2)$ is continuous. This gives the desired inequality.

Answer 2

(After some discussion in the comments, this is a proof to a different result.)

Lemma. A necessary and sufficient condition for a linear function $u$ to be continuous between two normed spaces $\mathrm{V}$ and $\mathrm{W}$ is that there should exists a $c > 0$ such that $\| u(x) \| \leq c \|x\|$ for all $x \in \mathrm{V}.$

Proof. The condition is trivially sufficient, let us focus on the necessity. Then $u$ is continuous at zero and this means that the inverse image by $u$ of the unit ball in $\mathrm{W}$ contains a ball $\mathrm{B} = \mathrm{B}(0; \frac{2}{c})$ in $\mathrm{V}.$ Then, for any $x \in \mathrm{V},$ we have $\dfrac{x}{c \|x\|} \in \mathrm{B}$ and so its image is inside the unit ball of $\mathrm{W},$ implying the desired result by linearity. $\square$

Lemma. If two norms $\| \cdot \|_1$ and $\| \cdot \|_2$ in a given vector space engender the same topology (i.e. they furnish the vector space with the same open sets), then there exists constants $C, c > 0$ such that $c \| \cdot \|_1 \leq \| \cdot \|_2 \leq C \| \cdot \|_1.$

Proof. Let $\mathrm{V}$ be the given vector space endowed with the norm $\| \cdot \|_1$ and let $\mathrm{W}$ the same vector space but now endowed with $\| \cdot \|_2.$ By the previous lemma, and upon noticing the hypothesis is equivalent to stating that the identity (linear) function is continuous $\mathrm{V} \to \mathrm{W}$ and $\mathrm{W} \to \mathrm{V}$, the result follows.

To your exercise. It is immediate from the previous lemma, in fact, the condition of completeness can be entirely dropped (as opposed to use Banach's open function theorem). What I showed is that any two norms that engender the same topology on a given vector space must necessarily be uniformly equivalent; futhermore, the same result and proof holds for Fréchet spaces.