Non equivalent norms I am doing some linear analysis problems and am faced with the following question
Conclude that if two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ on a (complex) vector space V are not equivalent, there exists a linear functional $f : V → \mathbb{C}$ which is continuous with respect to one of the two norms, and discontinuous with respect to the other. 
My idea was to use the contrapositive; we suppose that there are $a,b>0$ such that for all $f$ linear functional $V\to \mathbb{C}$  $$|f(x)|<a\|x\|_1$$ $$|f(x)|<a\|x\|_2$$
Then we look at the support functionals $f_x^{(i)}$ with respect to the $\|\cdot\|_i$ norm and plug them in to get
$$\|x\|_2<a\|x\|_1$$ $$\|x\|_1<b\|x\|_2$$
from which the result follows. But my problem is that here is that I am not used to constructing the support functional like this, that is getting two different support functionals on the same space, but with regard to different norms. Is this a valid method? If not, how is best to approach this problem.
 A: WLOG assume that $\not\exists M > 0$ such that $\|\cdot\|_2 \le M\|\cdot\|_1$. Suppose that for any linear functional $f : V \to \mathbb{C}$ holds $$f \text{ continuous w.r.t. } \|\cdot\|_2 \implies f \text{ continuous w.r.t. } \|\cdot\|_1$$
Note that if a sequence $(x_n)_n$ in $V$ converges weakly  w.r.t $\|\cdot\|_1$, then it also converges weakly w.r.t. $\|\cdot\|_2$:
\begin{align}
x_n \xrightarrow{w_1} x &\implies f(x_n) \xrightarrow{n\to\infty} f(x), \forall f \text{ linear functional continuous w.r.t. } \|\cdot\|_1 \\
&\implies f(x_n) \xrightarrow{n\to\infty} f(x), \forall f \text{ linear functional continuous w.r.t. } \|\cdot\|_2 \\
&\implies x_n \xrightarrow{w_2} x
\end{align}
We have $\|\cdot\|_2 \not\le n^2\|\cdot\|_1, \forall n \in \mathbb{N}$ so there exists a sequence $(x_n)_n$ in $V$ such that $\|x_n\|_1 = 1$ and $\|x_n\|_2 \ge n^2$ for all $n \in \mathbb{N}$.
We have
$$\left\|\frac1n x_n\right\|_1 = \frac1n \xrightarrow{n\to\infty} 0$$
so $$\frac1n x_n \xrightarrow{\|\cdot\|_1} 0 \implies \frac1n x_n \xrightarrow{w_1} 0 \implies \frac1n x_n \xrightarrow{w_2} 0$$
Weakly convergent subsequence in a normed space is also bounded so we conclude that $\left( \frac1n x_n\right)_n$ is bounded w.r.t $\|\cdot\|_2$.
However
$$\left\|\frac1n x_n\right\|_2 \ge \frac1n \cdot n^2 = n$$
which is a contradiction.
Therefore there exists a linear functional $f$ which is continuous w.r.t $\|\cdot\|_2$ but not w.r.t $\|\cdot\|_1$.
