Example of a sequence on an infinite-dimensional vector space with respect to different norms [closed]

I am stuck with the following problem:

Give example of an infinite dimensional vector space V and two norms $\theta$ and $\rho$ on V and the sequence $\{x_{n}\}_{n\geq 1}$ of V such that:

• The sequence $\{x_{n}\}_{n\geq 1}$ is cauchy in $(V, \rho)$
• The sequence $\{x_{n}\}_{n\geq 1}$ is not cauchy in $(V, \theta)$

It is easy to find a vector space, a norm, and a sequence that meet one of the items but I do not know how to proceed.

Any help is appreciated.

Solution 1:

Consider $V=\mathbb{R}[x]$, the space of real valued one variable polynomials. For each polynomial $f \in V$ suppose

$$||f||_{\theta}=\int_{[0,0.5]} |f(x)|dx \text{ and }||f||_{\rho}=\max(|f|)=\max\{|f(x)| ; x \in [0,1]\},$$

it is not hard to see that $\theta$ and $\rho$ are norms on $V$.

Now for the sequence $\{p_{n}\}$, where $p_{2n} = x^n$ and $p_{2n-1} = 0$, $p_{n}$ is Cauchy in $(V,\theta)$. Take $\epsilon > 0$ and $N$ such $\dfrac{2}{N} < \epsilon$, now for $n,m \geq N$ we have $||p_{n}-p_{m}||_{\theta} \leq \dfrac{1}{n}+\dfrac{1}{m} \leq \dfrac{2}{N} < \epsilon$; but it is not Cauchy in $V,\rho$ as $||x^{2n}-x^{2m-1}||_{\rho}=1$ for each $n,m$.

I think the point is to find the second norm, $\rho$, such that the the sequence you have found in the first part is not Cauchy.