Short answer. That means there exists an injection $\iota:C([0,1])\to L^\infty([0,1])$ and one can identify $C([0,1])$ with the image $\iota(C([0,1]))$, which is a subset of $L^\infty([0,1])$.

Now a long one.

We say that a measurable function $f: [0,1] \to {\mathbf C}$ is essentially bounded if there exists an M such that ${|f(x)| \leq M}$ for almost every $x\in[0,1]$, and define $\\\|f\\\|_{L^\infty}$ to be the least $M$ that serves as such a bound.

Let $\mathscr{L}^\infty([0,1])$ denote the set of measurable functions on $[0,1]$ that are essentially bounded and let $\sim$ be the equivalent relation on the set (two measurable functions are equivalent if they agree almost everywhere on $[0,1]$). Then $$L^\infty([0,1]) =\mathscr{L}^\infty([0,1])/\sim.$$ Here, we only care about the set structure, so the quotient is a "quotient set."

Clearly, $C([0,1])$ is a subset of $\mathscr{L}^\infty([0,1])$. So one can define the map $\iota:C([0,1])\to L^\infty([0,1])$ with $\iota(f)= [f]$ where $[f]$ denote an equivalent class in $L^\infty([0,1])$.

By continuity of functions in $C([0,1])$, one can show that the map $\iota$ is injective. In other words, there is a bijection between the set $C([0,1])$ and its image $\iota(C([0,1]))\subset L^\infty([0,1])$. Note that "bijection" is "set isomorphism." In this sense, one can say that $C([0,1])$ is a subset of $L^\infty([0,1])$.

With more structures on $L^\infty([0,1])$, say a normed vector space equipped with the norm topology, one can show that $C([0,1])$ is not dense in $L^\infty([0,1])$, which really means $\iota(C([0,1]))$ is not dense in $L^\infty([0,1])$.

Such identification is so common in analysis that people just use the short version.

posted 12 months ago

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5mo ago

Snoopy‭

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