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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).

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Errr... why does $\|\cdot\|_\ast = \|\cdot\|_C$ not fulfill the assumption? – shuhalo Jan 6 '12 at 16:41
@Martin: Because $f_n \to 0$ pointwise doesn't imply $f_n \to 0$ uniformly... – t.b. Jan 6 '12 at 16:44
up vote 2 down vote accepted

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.

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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 $\lim_{n\to\infty}\lVert f_n\rVert_C\to 0$....

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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. – NGY Jan 6 '12 at 17:26

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