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Let $(X,\Omega,\mu)$ be a measure space. If $1 < p < \infty,$ $f\in L_p$ and $\varepsilon >0$ then there exists a simple function $\phi$ where $\|f-\phi\|_p<\varepsilon$.

We did a problem similar to this one but when $p=1$. We had this particular question on homework a few weeks ago and we just had it again on an exam and I didn't get it.

Here is the direction I tried to go in:

If $f \in L_p$ and $\varepsilon > 0$ then we'll write $f=f^+-f^-$ where $f^+=\sup\{0,f\}$ and $f^-=\sup\{0,-f\}$. There then exists an increasing sequence of non-negative simple functions that converges to $f^+$, say $\phi_n \rightarrow f^+$ and similarly for $f^-$ say $\psi_n \rightarrow f^-$.

Goal: Find some simple function $\phi$ where $$\int \left\lvert f - \phi \right\rvert^p \mathrm{d}\mu < \varepsilon.$$

I propose some $\phi = \phi_n - \psi_n$ could be a canditate with appropriate $n \in \mathbb{N}$. So

$$\left\lvert f - \phi \right\rvert^p = \left\lvert f^+-f^- - (\phi_n-\psi_n) \right\rvert^p = \left\lvert (f^+-\phi_n)-(f^--\psi_n) \right\rvert^p.$$

At this point I am having problems. I don't think triangle inequality will hold in that sense that $$\left\lvert a-b \right\rvert^p \leq \left\lvert a \right\rvert^p+\left\lvert b \right\rvert^p$$ for any $a,b \in \mathbb{R}$. So I am thinking even if I could manage to do this, I would still have something like $$\int \left\lvert f-\phi \right\rvert^p \mathrm{d}\mu \leq \int \left\lvert f^+-\phi_n \right\rvert^p \mathrm{d}\mu + \int \left\lvert f^--\psi_n \right\rvert^p \mathrm{d}\mu$$ which doesn't seem to help if $\mu(X)=\infty$. Was thinking I could say something like: $$\int \left\lvert f^--\psi_n \right\rvert^p \mathrm{d}\mu < \int (\varepsilon^{1/p})^p \mathrm{d}\mu$$ but this falls apart if $\mu(X)=\infty$.

Any help is appreciated thanks.

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You're making things complicated by considering $f$ to be complex from the beginning. Start with the case $f \ge 0$. There exists a sequence $\{\phi_n\}$ of simple functions that converges to $f$ such that $$ 0 \le \phi_1 \le \phi_2 \le \cdots \le f. $$

Since both $\phi_n$ and $f$ are non-negative, we have $$ \left|f - \phi_n\right|^p \le f^p. $$

Since $f \in L^p(\mu)$, by the dominated convergence theorem $$ \lim_{n\to\infty} \int_X\left|f - \phi_n\right|^p \,d\mu = 0. $$

Thus, $\|f - \phi_n\|_p$ goes to $0$ as $n \to \infty$. It follows that we can find $N$ for which $\|f - \phi_N\|_p < \epsilon$ for any $\epsilon > 0$.


Now for the general case, write $f = (u^+ - u^-) + i(v^+ - v^-)$. Approximate each non-negative function on the RHS with a simple function, and use the triangle inequality to get the desired result.

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  • $\begingroup$ I appreciate it, I should be able to take it from here, if not I will come back and ask more questions. $\endgroup$ – Frudrururu May 9 '13 at 0:33
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For critique and for my own "records" of sorts, I will submit my solution here as an answer though I do accept Ayman's answer as the "accepted answer" for this question.

Lemma: If $f \geq 0$, $f \in L_p$, and $\varepsilon > 0$ then there exists a simple function $\phi$ where $\|f-\phi\|_p < \varepsilon$.

Proof: As $f \geq 0$ and measurable there exists an increasing non-negative sequence of simple function $\phi_n$ that converges to $f$. We have then $\lim_{n \rightarrow \infty}\phi_n=f$ or as $0 \leq f-\phi_n$, $$0=\lim_{n\rightarrow\infty}\left\lvert f-\phi_n \right\rvert.$$ Hence, $$0=\lim_{n\rightarrow\infty}\left\lvert f-\phi_n \right\rvert^p.$$ Since $0 \leq \phi_n \leq f$, $$\left\lvert f-\phi_n \right\rvert^p \leq \left\lvert f \right\rvert^p$$ and their integrals hold the same inequality and as $\left\lvert f \right\rvert^p \in L$, $\left\lvert f-\phi_n \right\rvert^p \in L$.

We then have by the dominated convergence theorem that $$\int \lim_{n\rightarrow\infty}\left\lvert f-\phi_n \right\rvert^p \mathrm{d}\mu = \lim_{n\rightarrow\infty} \int \left\lvert f-\phi_n \right\rvert^p \mathrm{d}\mu$$ or $$0=\lim_{n\rightarrow\infty} \int \left\lvert f-\phi_n \right\rvert^p \mathrm{d}\mu$$ which implies $$0=\lim_{n\rightarrow\infty} \left(\int \left\lvert f-\phi_n \right\rvert^p \mathrm{d}\mu \right)^{1/p}=\lim_{n\rightarrow\infty}\|f-\phi_n\|_p.$$ Given $\varepsilon >0$ then there exists some $N \in \mathbb{N}$ where $$\|f-\phi_N\|_p < \varepsilon.$$

Problem: Let $(X,\Omega,\mu)$ be a measure space. If $1 < p < \infty,$ $f\in L_p$ and $\varepsilon >0$ then there exists a simple function $\phi$ where $\|f-\phi\|_p<\varepsilon$.

Proof: Let $f \in L_p$ and $\varepsilon > 0$ be given. We have that $f=f^+-f^-$ and as $f \in L_p$,$f^+,f^- \in L_p$. Further as each $f^+,f^-$ is non-negative, there then exists simple functions $\phi$ and $\psi$ where $$\|f^+-\phi\|_p<\varepsilon/2 \text{ and } \|f^--\psi\|<\varepsilon/2.$$ Thus, $\theta=\phi-\psi$ a simple function has $$\|f-\theta\|_p=\|(f^+-\phi)-(f^--\psi)\|_p \leq \|f^+-\phi\|_p+\|f^--\psi\|<\varepsilon/2+\varepsilon/2=\varepsilon.$$

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