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

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.

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.