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Are simple functions dense in $L^\infty$? I've been able to show this for finite measure spaces but not in general.

Rudy the Reindeer
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Jamba
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4 Answers4

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If $f$ is bounded, then the function that has value $k\cdot\varepsilon$ on the set where $k\cdot\varepsilon\leq f(x)<(k+1)\cdot\varepsilon$ (for each $k\in\mathbb Z$) is a simple function whose $L^\infty$ distance to $f$ is at most $\varepsilon$.

Jonas Meyer
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  • What would happend if $f$ is unbounded? – user62089 Mar 17 '13 at 20:18
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    @pondy: I'm not sure exactly what you mean to ask. If $f$ is not essentially bounded then the equivalence class of functions a.e. equal to $f$ is not in $L^\infty$. Each element of $L^\infty$ has a bounded representative of its equivalence class. If, say, $f:[0,1]\to\mathbb R$ is an unbounded function, then for every simple function $g:[0,1]\to\mathbb R$, $f-g$ is also unbounded. – Jonas Meyer Mar 17 '13 at 20:39
  • I am sorry, that was a stupid doubt. Also I wanted to know if all functions in $L^p$ where $p < \infty$ are bounded. Or rather does the implication $\int |f|^p < \infty \Rightarrow |f| < \infty$ hold true. If not is there a counter example. – user62089 Mar 17 '13 at 20:55
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    @pondy: No. Simple counterexample: $1/\sqrt x$ on $(0,1)$. Basic idea: Function gets big but the sets on which it is big get smaller faster. – Jonas Meyer Mar 17 '13 at 21:08
  • @pondy: I just remembered the following, related to the question in your last comment: http://math.stackexchange.com/questions/90668/a-function-that-is-lp-for-all-p-but-is-not-l-infty – Jonas Meyer Mar 17 '13 at 21:23
  • Are the integrable simple functions too dense in $\mathcal{L}^{\infty}$. – user62089 Mar 18 '13 at 04:57
  • @pondy: $L^\infty$ functions on what measure space? If the space has finite measure, obviously yes. If the space has infinite measure, obviously no. – Jonas Meyer Mar 18 '13 at 05:00
  • I don't see why the finiteness assumption matters. Can you elaborate a little. – user62089 Mar 18 '13 at 05:02
  • @pondy: Do you have any thoughts on the problem? Where did you get stuck? Under what conditions is a simple function integrable? I do not want to elaborate on this hint here, now; you are of course welcome to ask this as a new question if you get stuck. – Jonas Meyer Mar 18 '13 at 05:04
  • A simple function is integrable if it has a finite support. Is this the only reason or is there more to it? And I see why it is true - DCT. – user62089 Mar 18 '13 at 05:07
  • @pondy: So on a measure space with finite measure, which simple functions are integrable? On a measure space with infinite measure, can you find a bounded function that is not close to any integrable simple function? – Jonas Meyer Mar 18 '13 at 05:09
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By simple approximation theorem(see Real analysis by Royden and Fitzpatrick), if $|f| \leq M$ then there exists a sequence of simple functions $\phi_{\epsilon}, \psi_{\epsilon}$ such that \begin{eqnarray} \phi_{\epsilon}(x)\leq f(x) \leq \psi_{\epsilon}(x) \quad \text{for all } x \in \mathbb{R} \end{eqnarray} such that $\psi_{\epsilon}(x)-\phi_{\epsilon}(x) \leq \epsilon.$

Now, if $g\in L^{\infty}(\mathbb{R})$ then there exists $\tilde{g}$ such that $g=\tilde{g}$ a.e. and $|\tilde{g}| \leq M$ for some $M>0.$ Thus applying the simple approximation theorem $\phi_{\epsilon}$ converges to $\tilde{g}$ in $L^{\infty}$ which further implies $\phi_{\epsilon}$ converges to ${g}$ in $L^{\infty}$

Veronica
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Consider the function $f \equiv 1$ on $L^{\infty}(\mathbb{R},\mu)$, then $\|s-f\|_{\infty} > \varepsilon$ for any simple function $s$. In particular we can have $\|s-f\|_{\infty} > 1$. So simple functions may not be dense in $L^{\infty}$. But note that here I have assumed that $ \mu \big(\{x \in X : s(x) \neq 0\}\big) < \infty $.

Himanshu
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From definition of $L_{\infty}(X,\mathbb{X},\mu)$ as the set of all function essentially bounded , where $f:X\longrightarrow \mathbb{R}$ a function $\mathbb{X}$-measurable is essentially bounded iff there is a bounded function $g:X\longrightarrow \mathbb{R}$ such that $g=f$ on $\mu$-a.e, we have that for any $f\in L_{\infty}$ it can be found a representative $g$ from equivalence class $f$ that is bounded.

So, for that function we can find a sequence $(\phi_n)$ of simple function that converges uniformly to $g$ in $L_{\infty}$, so that sequence converges also to $f$ in $L_{\infty}$.

As all simple function belongs to $L_{\infty}$, this convergency is given in $L_{\infty}$, so $\parallel f-\phi_n \parallel_{\infty} \rightarrow 0$

Irene
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    This answer doesn't prove anything, because its second paragraph is just a restatement of the question without any proof. – Alex M. May 01 '20 at 18:34