4
$\begingroup$

Question:

Let $X=C[0,1]$, show that there is no such norm $\lVert\cdot\rVert_*$ on $X$ that for any series $\{f_n\}_{n=1}^{\infty}\subset X$,

$$\lim_{n\to\infty}\lVert f_n\rVert_*\to 0\Longleftrightarrow \lim_{n\to\infty}f_n(t)=0,\quad\forall t\in[0,1]$$

============

I've tried to define a new norm (supposing such $\lVert\cdot\rVert_*$ exists): $$\lVert f\rVert_+=\lVert f\rVert_*+\max_{t\in[0,1]}|f(t)|=\lVert f\rVert_*+\lVert f\rVert_C$$

it is easy to show that $\lVert\cdot\rVert_+$ is a complete norm on $X$ (so is $\lVert\cdot\rVert_C$), and this implies $\lVert\cdot\rVert_+$ and $\lVert\cdot\rVert_C$ are equivalent norms, so there is a constant $M$ s.t.

$$\lVert f\rVert_*\leq M\lVert f\rVert_C,\quad f\in X$$

and I got stuck at the above inequality (or maybe it is useless).

$\endgroup$
  • $\begingroup$ Errr... why does $\|\cdot\|_\ast = \|\cdot\|_C$ not fulfill the assumption? $\endgroup$ – shuhalo Jan 6 '12 at 16:41
  • $\begingroup$ @Martin: Because $f_n \to 0$ pointwise doesn't imply $f_n \to 0$ uniformly... $\endgroup$ – t.b. Jan 6 '12 at 16:44
  • 1
    $\begingroup$ See also math.stackexchange.com/questions/33476/… $\endgroup$ – user940 Jan 6 '12 at 22:42
3
$\begingroup$

In fact, the topology pointwise convergence of continuous functions on $[0,1]$ is not metrizable. Indeed, denote for a continuous function $f\colon [0,1]\to\mathbb R$, $J$ a finite subset of $[0,1]$ and $\varepsilon>0$ $$B(f,J,\varepsilon)=\left\{g\colon [0,1]\to\mathbb R,g\mbox{ continuous }\mid \forall t\in J, |g(t)-f(t)|<\varepsilon\right\}.$$ It's a basis for the topology of the uniform convergence. Suppose that the null function $\mathbf 0$ has countable basis of neighborhood $\{V_n\}$. Then for each $n$, we can find a finite subset $J_n$ of $[0,1]$ and $\varepsilon>0$ such that $B(\mathbf 0,J_n,\varepsilon_n)\subset V_n$. Since $[0,1]$ is uncountable, let $j\in [0,1]\setminus \bigcup_n J_n$. Put $V:=B(\mathbf 0,\{j\},1)$. We can find a continuous function $g_n\colon [0,1]\to\mathbb R$ which is in $B(0,J_n,\varepsilon_n)$ (take a function which is $0$ on $J_n$ (which is finite) and $g_n(j)=2$. Hence $g_n\in B(\mathbf 0,J_n,\varepsilon_n)\setminus B(\mathbf 0,\{j\},1)$ and $B(\mathbf 0,J_n,\varepsilon_n)$ is never contained in $V$, which contradicts the fact $\{V_n\}$ is a basis of neighborhood of this topology. In particular, this one is not metrizable.

$\endgroup$
3
$\begingroup$

You also get

$$\lVert f\rVert_C \leq M_1 \lVert f\rVert_* ,\quad f\in X \,$$

for some $M_1$. This implies that if $\lim_{n\to\infty}f_n(t)=0,\quad\forall t\in[0,1]$ then $\lim_{n\to\infty}\lVert f_n\rVert_C\to 0$....

$\endgroup$
  • 1
    $\begingroup$ How to prove it(the inequality containing $M_1$)? I don't think $\lVert\cdot\rVert_+$ and $\lVert\cdot\rVert_C$ are equivalent norms implies that. $\endgroup$ – NGY Jan 6 '12 at 17:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.