Show that $(X,\|\cdot\|)$ is complete if and only if $(X,|\!|\!|\cdot|\!|\!|)$ is complete when both norms are equivalent. Problem: "Suppose that on the same normed linear space, $X$, you have two equivalent norms, $\|\cdot\|$ and $|\!|\!|\cdot|\!|\!|$. Show that $(X,\|\cdot\|)$ is complete if and only if $(X,|\!|\!|\cdot|\!|\!|)$ is complete."
My attempt: Recall that if two norms are equivalent then $\exists\, a,b\in\mathbb R^+$ such that $a\|x\|\le|\!|\!|x|\!|\!|\le b\|x\|,\,\forall x\in X$. Also recall that a normed linear space is complete if every Cauchy sequence of elements converges to it's limit in the space.
Suppose, then, that $(X,\|\cdot\|)$ is complete and consider a Cauchy sequence of elements in $X$, $(x_n)$, which converges to it's limit $x\in X$. Then $\|x_n-x\|\to0$ as $n\to\infty.$ But then we have that,
$$a\|x_n-x\|\le|\!|\!|x_n-x|\!|\!|\le b\|x_n-x\|,\,\forall n\in\mathbb N$$
$$\implies0\le|\!|\!|x_n-x|\!|\!|\le0,\,\forall n\in\mathbb N$$
Which means thats $(x_n)$ is a convergent sequence in $X$ under $|\!|\!|\cdot|\!|\!|$. But we know in a metric space that convergent sequences are necessarily Cauchy. Hence this Cauchy sequence converges to it's limit in $X$ under $|\!|\!|\cdot|\!|\!|$. This $(X,|\!|\!|\cdot|\!|\!|)$ is complete. A completely analogous argument holds for the other implication.
Is my attempt and reasoning correct? My uncertainty stems from the solutions I have to this problem from my class. There, the lecturer has shown seperately that $(x_n)$ converges in one norm iff it converges in the other norm, and that $(x_n)$ is Cauchy in one norm iff it is Cauchy in the other. Is it not sufficient to merely show the latter?
 A: I think the reasoning is not good enough. Actually one should assume first that $(x_{n})$ is Cauchy in $|\|\cdot\||$ and $\|\cdot\|$ is complete. Then there is some $\|x_{n}-x_{m}\|\leq C|\|x_{n}-x_{m}\||\rightarrow 0$, so $(x_{n})$ is Cauchy in $\|\cdot\|$. Then there is some $x\in X$ such that $\|x_{n}-x\|\rightarrow 0$. Since $|\|x_{n}-x\||\leq D\|x_{n}-x\|\rightarrow 0$, so $(x_{n})$ converges to $x$ in $|\|\cdot\||$.
And by symmetry one does in the other way as well.
A: Assume $m\|\cdot\| \le |\!|\!|\cdot|\!|\!| \le M\|\cdot\|$ for some constants $m,M > 0$.
Take a Cauchy sequence $(x_n)_{n=1}^\infty$ in $\|\cdot\|$.
Then we have
$$|\!|\!|x_k - x_n|\!|\!| \le M\|x_k - x_n\| \xrightarrow{k,n\to\infty} 0$$
so $(x_n)_{n=1}^\infty$ is also Cauchy in $|\!|\!|\cdot|\!|\!|$.
Similarly, take a Cauchy sequence $(x_n)_{n=1}^\infty$ in $|\!|\!|\cdot|\!|\!|$.
Then we have
$$\|x_k-x_n\| \le \frac1m|\!|\!|x_k - x_n|\!|\!| \xrightarrow{k,n\to\infty} 0$$
so $(x_n)_{n=1}^\infty$ is also Cauchy in $\|\cdot\|$.
Exactly the same reasoning shows that $(x_n)_{n=1}^\infty$ converges in $\|\cdot\|$ to $x \in X$ if and only if it converges in $|\!|\!|\cdot|\!|\!|$ to  the same $x$. Therefore, $\|\cdot\|$ and $|\!|\!|\cdot|\!|\!|$ have the same Cauchy sequences and the same convergent sequences.
Therefore, the implication $$\text{ Cauchy } \implies \text{ convergent }$$ is true for $\|\cdot\|$ if and only if it is true for $|\!|\!|\cdot|\!|\!|$.

The converse also holds, sort of.
Lemma:

Let $\|\cdot\|$ and $|\!|\!|\cdot|\!|\!|$ be two norms on $X$ such that for all sequences $(x_n)_{n=1}^\infty$ in $X$ we have:
$$(x_n)_{n=1}^\infty \text{ Cauchy in } \|\cdot\| \implies (x_n)_{n=1}^\infty \text{ bounded in } |\!|\!|\cdot|\!|\!|$$
Then there exists $M > 0$ such that $|\!|\!|\cdot|\!|\!| \le M\|\cdot\|$.

Proof: Assume there does not exist such a constant $M$. Then you can construct a sequence $(x_n)_{n=1}^\infty$ such that $\|x_n\| = 1$ and $|\!|\!|x_n|\!|\!| > n^2$ for all $n \in \mathbb{N}$. Then $\left\|\frac1n x_n\right\| = \frac1n \xrightarrow{n\to\infty} 0$ so $(x_n)_{n=1}^\infty$ is convergent (so in particular Cauchy) in $\|\cdot\|$. Therefore $\left(\frac1n x_n\right)_{n=1}^\infty$ must be bounded in $|\!|\!|\cdot|\!|\!|$ which is a contradiction.
As a consequence, we get that if $\|\cdot\|$ and $|\!|\!|\cdot|\!|\!|$ have the same Cauchy sequences (or the same convergent sequences), they are necessarily equivalent.
A: Here's a direct way to demonstrate this, by using preservation of the property under certain mappings. 
Let $ \left(E, \| \|_{1} \right)$ and $ \left(E, \| \|_{2} \right)$ two $\mathbb{K}$ finite dimensionial vector space. 
Consider the following application:
\begin{aligned}
 \left(E, \| \|_{1} \right) & \rightarrow \left(E, \| \|_{2} \right)  \\
x  &  \mapsto  x 
\end{aligned}
which is a linear lipschitz (using the equivalence of the two norms) continuous bijection (thus, a homeomorphism).
The pre-image of a complete space by a uniformly continuous and bijective function being complete it follows that $\left(E, \| \|_{1} \right)$ is complete if and only if $\left(E, \| \|_{2} \right)$ is complete.
