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I know that the continuous functions $f:(0,\infty) \rightarrow R $ are dense in $L_p(0,\infty)$, with respect to the norm $|| \space||_p$. Therefore, if $f\in L_p(0,\infty)$ then there exists a sequence of continous functions $\{f_n\}$ in $L_p$ such that $f_n \rightarrow f$. I'm wondering if there exists a sequence that does this, but also is non-decreasing, meaning $f_{n+1}\geq f_n$ pointwise for each $n$. I believe this to be true, but I haven't been able to prove it. So, is this true? If it is, I would appreciate any tips on how to prove it.

Thanks!

user111
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MathUser123
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  • @KaviRamaMurthy, the pointwise infimum of countably many continuous functions need not be continuous. For example, take $g_n(x) = x^n$ on $[0,1]$. If we let $f_n(x) = \inf{g_n, g_{n+1}, \ldots}$, then $f$ is discontinuous at $x=1$. I suspect that OP's claim is false based on this post. – Liam Apr 30 '19 at 00:26

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I suspect that this is not true. Inspired by this post here, consider a fat Cantor subset, $C$, of $[0,1]$ (a nowhere dense subset of $[0,1]$ with positive measure) and its indicator function $\chi_C$. Let $f$ be a continuous function with $0\leq f \leq \chi_C$. Let $x$ be any point in $C$. Since $C$ is nowhere dense, for any natural $k>0$, the interval $(x-\frac{1}{k}, x+\frac{1}{k})$ is not fully contained in $C$, so there is some $x_k$ in this interval with $\chi_C(x_k) = 0$. Since $f$ is bounded above by $\chi_C$, we must also have $f(x_k) = 0$. We can choose these $x_k$'s so that $x_k\to x$ as $k\to \infty$. By the continuity of $f$ we must then have $f(x) = 0$.

So any continuous function bounded above by $\chi_C$ must be identically zero. But then there is no hope for us to find a sequence of continuous functions that increase to $\chi_C$ that approximate $\chi_C$ in any $L^p$ space since $C$ has positive measure.

Liam
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This is not true. Assume, on the contrary, that there exists a sequence of continuous functions $f_n\uparrow f$ in $L^p$. Then, there would be a subsequence, that we still denote by $f_n$ such that $f_n\uparrow f$ almost everywhere.

Then, following this answer (for completeness we reproduce it), let $\lambda$ be the Lebesgue measure, and let $M$ be any measurable set s.t. for any interval $I$, $$\lambda(I) > \lambda(M \cap I) > 0,$$ (see here to construct it). Let $f(x) = \mathbb 1_M(x)$ be the characteristic function of $M$.

Let $x_0\in M$ s.t. $f_n(x) \to f(x)$ (such point exists, because $f_n \to f$ a.e., and $M$ has positive measure). Then, for some $N$, we have $f_N(x_0) \geq 1/2$, and, as $f$ is continuous, for some $\epsilon > 0$ we have $$f_N(x) \geq 1/4,\qquad x \in (x_0 - \epsilon, x_0 + \epsilon),$$ which implies that $$\forall n\geq N,\quad f_n(x) \geq 1/4,\qquad x \in (x_0 - \epsilon, x_0 + \epsilon).$$ But that means that $f_n$ doesn't converge to $0$ on $N = (x_0 - \epsilon, x_0 + \epsilon) \setminus M$, which has positive measure, while $f$ is $0$ on this set.

user111
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