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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.

mechanodroid
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Elie Bergman
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1 Answers1

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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 $\|\cdot\|_1$.

mechanodroid
  • 46,490