Supremum Metric and L1 Metric on Closed Real Intervals are not Topologically Equivalent

Theorem

Let $S$ be the set of all real functions which are continuous on the closed interval $\closedint a b$.

Let $d_1$ be the $L^1$ metric on $S$:

$\ds \forall f, g \in S: \map {d_1} {f, g} := \int_a^b \size {\map f t - \map g t} \rd t$

Let $d_\infty$ be the supremum metric on $S$:

$\ds \forall f, g \in S: \map {d_\infty} {f, g} := \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}$

Then $d_1$ and $d_\infty$ are not topologically equivalent.

Proof

Let $f, g \in S$.

Then by definition of supremum metric:

$\forall x \in \closedint a b: \size {\map f x - \map g x} \le \map {d_\infty} {f, g}$

Hence by ...

$\map {d_1} {f, g} = \ds \int_a^b \size {\map f x - \map g x} \rd x \le \paren {b - a} \map {d_\infty} {f, g}$

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and so:

$\map {B_\epsilon} {f; d_\infty} \subseteq \map {b_{\paren {b - a} } } {f; d_1}$

Now let $f_0$ denote the constant mapping:

$\forall x: \map {f_0} x = 0$

We show that $\map {B_1} {f_0; d_\infty}$ is not $d_1$-open.

Aiming for a contradiction, suppose instead that $\map {B_1} {f_0; d_\infty}$ is $d_1$-open.

Then we should have:

$\map {B_1} {f_0; d_1} \subseteq \map {B_1} {f_0; d_\infty}$

for some $\epsilon \in \R_{>0}$.

But for $\epsilon > 0$ there exists a continuous function on $\closedint a b$ such that:

$\map {d_1} {f, f_0} < \epsilon$

yet:

$\map {d_\infty} {f, f_0} = 1$

So:

$f \in \map {B_1} {f_0; d_1}$

but:

$f \notin \map {B_1} {f_0; d_\infty}$

which contradicts our deduction that $\map {B_1} {f_0; d_1} \subseteq \map {B_1} {f_0; d_\infty}$.

Hence it cannot be the case that $\map {B_1} {f_0; d_\infty}$ is $d_1$-open.

The result follows.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics: Example $2.4.6$