Functional Analysis, topology induced by infinite norm

Question

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

Answer 1

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.

Answer 2

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 $\varepsilon$, $B_{L^1}(0, \varepsilon) \subsetneq B_{\|\cdot\|_\infty}(0,\varepsilon)$.

Answer 3

Define $F:C^1[0,1]\to \Bbb{R}, f\mapsto f(1)$. Every uniform convergent (that is, convergent by the sup-norm) function sequence $f_n\to f$ has $f_n(1) \to f(1)$, however the sequence $f_n=x^n$ converges in $L_1$ to $f=0$, but $f_n (1)=1$ for all $n$. This shows that $F$ is (sequentially) continuous w.r.t. to the sup-norm but not w.r.t. the $L_1$ norm, which means that the $L_1$ norm is coarser.