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Trying to understand enter image description here

For (b), I understand that there are two claims are proposed in respective the cases $[1,\infty]$ and $[1,\infty)$.

i. For $p\in[1,\infty]$, one has $$\|\rho_k * f\|_p\le\|f\|_p,\qquad (i.e.\quad \rho_k * f\in L^p(\mathbb{R})\quad)$$ ii. For $p\in[1,\infty)$, one has $$\|\rho_k * f-f\|_p\to 0,\qquad (k\to\infty)$$ (convergence in $p-norm$)

However, I have a rather vague idea of the difference of the two, and I cannot explain to myself the meaning/implication of the two equations and how they differ. 1) Could anyone describe to me the meaning of the two equations, the significance of including $\infty$, in words/simple mathematical language?

2) In the case $p=\infty$, equation in $(ii)$ no longer holds, why? And equation in $(ii)$ always implies equation in $(i)$, right?

Giovanni
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math101
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1 Answers1

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Notice that $(i)$ is not only saying that $\rho_k * f \in L^p$. The information encoded in the inequality is that the convolution with one of the $\rho_k$ does not increase the $L^p$-norm of $f$. This is true for every $p$, including $\infty$, and for every $k$.

On the other hand $(ii)$ exploits the fact that $\rho_k$ is a $\delta$-sequence to recover $f$ when $k$ approaches $\infty$. One of the main differences with $(i)$ is that it is a statement about the limit as $k$ goes to infinity, so there is no information about "small" values of $k$. Hence it is not correct to say that $(ii)$ implies $(i)$.

Moreover, the convergence in $(ii)$ implies that for $k$ large $\rho_k * f$ is close to $f$ in an $L^p$ sense, but there is no obvious inequality between the norms of these two $L^p$ functions.

This is important: the reason why $p = \infty$ is not included in $(ii)$ is that the corresponding result is false. Indeed, assume by contradiction that the result is true, i.e. $$\|\rho_k * f - f\|_{\infty} \to 0. \tag 1$$ Notice that $\rho_k * f$ is continuous since $\rho_k$ is assumed to be continuous. Then by $(1)$ we have that the continuous sequence of functions $\{\rho_k * f\}$ that converges uniformly to $f$. This would imply that $f$ is continuous, hence yielding a contradiction (not every $L^{\infty}$ function is continuous!).

The following is a function in $L^{\infty}(\mathbb{R})$ that cannot be made into a continuous function by redefining it on a null a set.

Consider $\chi(x) = \chi_{[0,1]}(x)$ and assume by contradiction that $f$ is a continuous function such that $f(x) = \chi(x)$ almost everywhere. Say that the set where they differ is $E$. Let $x_0$ be the smallest point such that $f(x_0) = \chi(x_0) = 1$. Consider intervals of the form $(x_0 - 1/n,x_0)$: since they all have positive measure we can find $x_n \in (x_0 - 1/n,x_0) \cap E^c$. By definition of $x_0$ we can also assume that $\chi(x_n) = 0$, indeed $x_0$ is only a set of measure $0$ apart from $0$. Then $x_n \to x_0$, so by continuity of $f$ we also have $f(x_n) \to 0$. This gives a contradiction because $$0 = \lim_n 0 = \lim_n \chi(x_n) = \lim_n f(x_n) = f(x_0) = 1.$$

Giovanni
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  • I need some time to read your answer. Thanks. – math101 Oct 28 '15 at 10:09
  • I do not understand what you mean by "increase the $L^p$-norm. – math101 Oct 28 '15 at 10:37
  • No problem, take all the time you need! :) What I mean is just that item $(i)$ not only verifies that $\rho_k * f \in L^p$, it also gives an explicit and perhaps useful estimate in terms of the norm of $|f|_{L^p}$. – Giovanni Oct 28 '15 at 14:51
  • Thanks for your patience. I just need to get through some even more before I can understand your answer fully. For instance http://math.stackexchange.com/questions/1502890/evans-pde-p-714-change-of-variable-and-change-of-integration-region – math101 Oct 29 '15 at 03:51
  • I think I have trouble convergence in $p$ norm in general, i.e. $g^{\epsilon}\to g$ in $L^p$, and how it distinct from proving $| g^{\epsilon}|{L^p}\le |g|{L^p}$. Need some time to settle down. – math101 Oct 29 '15 at 03:57
  • $g^{\epsilon} \to g$ in $L^p$ means that $|g^{\epsilon} - g|_{L^p} \to 0$. The second inequality you wrote only says that the norm of $g^{\epsilon}$ is smaller or equal to the norm of $g$, but it does not say anything about how close it is, for example. – Giovanni Oct 29 '15 at 04:07
  • yes the inequality says that the lhs is controlled by $f$ in $L^p$. – math101 Oct 29 '15 at 04:11
  • i think the proof of the proposition I stated is pretty much the same as the proof of properties of mollifier in the appendix of Evans pde...right? – math101 Oct 29 '15 at 09:56
  • You are definitely correct, the ideas and the tricks you use are very similar if not exactly the same! – Giovanni Oct 29 '15 at 14:43
  • on your comment "not every $L^{\infty}$ function is continuous$, why? any examples – math101 Oct 29 '15 at 22:35
  • A remark first: the uniform convergence in my answer is actually uniform convergence on a set of full measure (i.e. outside a set of measure $0$). For an example of a discontinuous function that is in $L^{\infty}$ consider the characteristic function of the interval $[0,1]$ in $\mathbb{R}$. Notice that this function is really discontinuous: redefining it on a set of zero measure won't make it continuous. This function is obviously in $L^{\infty}$ since its essential supremum is $1$. – Giovanni Oct 29 '15 at 23:27
  • "Redefine it on a set of zero", could you show me mathematically how one would do this? – math101 Oct 30 '15 at 10:16
  • Take a function $f$ and consider a set of zero measure in the domain of $f$. Now define a new function $g$ that agrees with $f$ everywhere except on the chosen set of zero measure. You can define $g$ to be whatever you want there. For a more concrete example take the function $h(x) = 1$ for every $x \neq 0$ and $h(0) = 7$. This function is not continuous at $0$, but can be made into a continuous function by redefining its value at $0$ to be $1$. We changed $f$ at a point, which is a subset of $\mathbb{R}$ of zero measure. – Giovanni Oct 30 '15 at 14:48
  • In the 2nd last sentence of your answer, How do you know the convergence is uniform? not pointwise – math101 Nov 04 '15 at 05:15
  • http://math.stackexchange.com/questions/83519/convergence-in-l-infty-is-nearly-uniform-convergence – Giovanni Nov 04 '15 at 05:18
  • I don't quite get the proof, never saw this in a book. Would be appreciated if you could add a proof for that claim in the answer you provided – math101 Nov 04 '15 at 05:35
  • Step1: write down the definition of uniform convergence. Step2: write down the definition of convergence in $L^{\infty}$ Step3: they are the same outside a set of measure $0$. – Giovanni Nov 04 '15 at 05:37
  • yes it is clear. – math101 Nov 04 '15 at 06:15
  • Also, looking at the concrete example you provided. The example showed to me redefining h(x) on a set of measure 0 does turn the discontinuous function continuous. But before that you remarked that "redefining it on a set of zero measure won't make it continuous" Could you please clarify, your example did turn a discontinuous function $h(x)$ into a continuous function, which a bit inconsistent with your remark... – math101 Nov 04 '15 at 07:33
  • It is not inconsistent with the remark, it simply answers a different question. It extremely easy to come up with a function that cannot be made into a continuous function by redefining it on a null set. For instance consider the characteristic function of an interval on the real line. – Giovanni Nov 04 '15 at 14:07
  • Okay. Your remark tells the fact that, in some cases, we cannot remove the discontinuity even by redefining it on a null set. Then, could you please provide a concrete example according to this fact? (May be just expand your last sentence a bit...) – math101 Nov 04 '15 at 23:50
  • I have edited the answer to include a proof of this fact – Giovanni Nov 05 '15 at 00:37
  • I am looking at "Consider intervals of the form $(x_0 - 1/n,x_0)$: since they all have positive measure we can find $x_n \in (x_0 - 1/n,x_0) \cap E^c$." Here I see two claims. First, all points in the interval have positive measure, why? Second, how does the first claim enable us to find such a $x_n$ which is in $(x_0 - 1/n,x_0) \cap E^c$? – math101 Nov 05 '15 at 04:17
  • with they I am referring to the intervals, not the points. Indeed, the $n$-th interval has measure (length) $1/n$. Since $E$ has measure $0$ it cannot cover the entire interval, hence there is always an $x_n$ that belongs to $E^c$. – Giovanni Nov 05 '15 at 05:23
  • @ Giovanni : Regarding the counter example you provided for my question " $L^{\infty}(\mathbb{R})$ that cannot be made into a continuous function by redefining on a null set. " Do you have an example for $\mathbb{R}^n$? – math101 Dec 05 '15 at 03:09
  • Take the characteristic function of the unit cube – Giovanni Dec 05 '15 at 22:29