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In the $C[0,1]$ function space, do all norms induce a finer topology than the topology induced by infinite norm ($\sup$ norm)?

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Dear Louis, I edited your question. Can you check that I didn't change the intended meaning? Thank you. – Rudy the Reindeer Sep 19 '12 at 12:42
up vote 2 down vote accepted

To expand on Nate's answer:

To see that the $L^1$-topology is coarser than the $\sup$-topology it is enough to exhibit a set that is open in the $\sup$-topology but not in the $L^1$-topology. To this end, consider the open unit ball around zero in the $\sup$ norm, $B_{\|\cdot\|_\infty}(0, 1)$. If this ball was open in the $L^1$-topology we'd be able to find an $\varepsilon$ for every point $x$ in it such that the $\varepsilon$-ball $B_{L^1}(x, \varepsilon)$ was contained in $B_{\|\cdot\|_\infty}(0, 1)$.

To see that this is not true, consider the point $0$. Let $\varepsilon = \frac{1}{n}> 0$ be arbitrary and consider $B_{L^1}(0, \frac{1}{n})$. Define $$ f_n(x) = \begin{cases} 10 \sin(n \cdot 10 x) & x \in [0, \frac{\pi}{n \cdot 10}] \\ 0 & \text{ otherwise } \end{cases}$$

$f_n$ is continuous and $\int_{[0,1]} f_n dx = \frac{2}{n}$ but $\|f_n \|_\infty = 10$. Hence for every $\varepsilon$ we can find $n$ such that $f_n \in B_{L^1}(0, \varepsilon)$ but $f_n \notin B_{\|\cdot\|_\infty}(0,\varepsilon)$, hence for all $\varepsilon$, $B_{L^1}(0, \varepsilon) \subsetneq B_{\|\cdot\|_\infty}(0,\varepsilon)$.

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Certainly not. For example, the $L^1$ norm $\|f\|_{L^1} := \int_0^1 |f(x)|\,dx$ induces a coarser topology than the sup norm.

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Please include the absolute value within the integral (it is too short for an edit). – Vobo Sep 19 '12 at 12:16

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