convolution is uniformly continuous

\nonumber\]. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. Stack Exchange network consists of 182 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. We first make the observation that if $f: D \rightarrow \mathbb{R}$ is uniformly continuous on $D$ and $A \subset D$, then $f$ is uniformly continuous on $A$. That's perfect! \limsup_{\epsilon\to 0}\lVert(f*j_\epsilon)(\cdot)-f(\cdot)\rVert_\infty\leq \lVert j\rVert_1\cdot \sup_{x\in\mathbb{R},|y|<\delta}|f(x)-f(x-y)|, Keeping DNA sequence after changing FASTA header on command line. $$|f\star K_n(x)-f\star K(x)|\leqslant \left|\int_\mathbb Rf(x-t)K_n(t)\mathrm dt-\int_\mathbb Rf(x-t)K(t)\mathrm dt\right|\leqslant\lVert f\rVert_\infty\lVert K_n-K\rVert_1.$$. Let us show that $f$ is Hlder continuous on $D$. How does "safely" function in "a daydream safely beyond human possibility"? Give an example of a subset $D$ of $\mathbb{R}$ and uniformly continuous functions $f, g: D \rightarrow \mathbb{R}$ such that $fg$ is not uniformly continuous on $D$. Learn more about Stack Overflow the company, and our products. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Let $\varepsilon = 2$ and $\delta > 0$. We can now invoke Theorem 3.2.2 to conclude $\lim _{x \rightarrow a^{+}} f(x)$ exists. $$ The second term can be estimated as Assume f(x) is uniformly continuous for all x R and zero outside a bounded set. Is there a lack of precision in the general form of writing an ellipse? PDF 6 Jointly continuous random variables - University of Arizona Convolutions - Statlect This is clear for indicator functions of measurable sets of finite measure, hence for simple functions by linearity and thus for general $L^q$ functions with the usual approximation argument. Therefore convolution is a continuous bilinear map from $L^2([0,2\pi])$ to the space of bounded functions. Set $\delta=\delta_{0} / 2$. Is the function $F(w) = \int_{\mathbb{R}^2}g(z)f(w-z)dz$ continuous? Now, if we make $f$ uniform continuous, how can we ensure uniform convergence throughout $\mathbb{R}^n$? The conv result should sum y (t-3)-y (t-5) but it gives me: Theme Copy y=@ (t) 1.0* (t>=0). We will show that $f$ is uniformly continuous on $\mathbb{R}$. \end{cases}$. skinny inner tube for 650b (38-584) tire? In CP/M, how did a program know when to load a particular overlay? \lim _{x \rightarrow b^{-}} f(x), & \text { if } x=b \text {.} Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be given by $f(x)=\frac{x^{2}}{x^{2}+1}$. (C) for every two sequences $\left\{u_{n}\right\}$, $\left\{v_{n}\right\}$ in $D$ such that $\lim _{n \rightarrow \infty}\left(u_{n}-v_{n}\right)=0$, it follows that $\lim _{n \rightarrow \infty}\left(f\left(u_{n}\right)-f\left(v_{n}\right)\right)=0$. The following theorem shows one important case in which continuity implies uniform continuity. The convolution/sum of probability distributions arises in probability theory and statistics as the operation in terms of probability distributions that corresponds to the addition of independent random variables and, by extension, to forming linear combinations of random variables. Hlder condition - Wikipedia This follows by noting that if $|f(u)-f(v)|<\varepsilon$ whenever $u,v \in D$ with $|u-v|<\delta$, then we also have $|f(u)-f(v)|<\varepsilon$ when we restrict $u,v$ to be in $A$. The best answers are voted up and rise to the top, Not the answer you're looking for? I can see that convolution of two continuous functions is continuous but I'm not sure how to use the uniformity of limits. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. $$ Connect and share knowledge within a single location that is structured and easy to search. We will show first that $\lim _{x \rightarrow a^{+}} f(x)$ exists. Prove that each of the following functions is uniformly continuous on the given domain: Add texts here. For any > 0, the condition implies the function is uniformly continuous. I know we define the density of Z, fz as the convolution of fx and fy but I have no idea why to evaluate the convolution integral, we consider the intervals [0,z] and [1,z-1]. Here's a very similar result in my real analysis book: Theorem: Let $\{\phi_k\}$ be a sequence of functions in $L^1(\mathbb{R}^n)$ such that. prove that a function $f$ is uniformly continuous if and only if there exists a modulus of continuity for $f$, Showing a convolution is uniformly continuous, proof check, Showing a function is uniformly continuous, Arbitrarily close uniformly continuous functions, Non-persons in a world of machine and biologically integrated intelligences. The proof is now complete. Example 3.5.4 Let f: R R be given by f(x) = x2 x2 + 1. Uniform continuity and translation invariance, Statement from SO: June 5, 2023 Moderator Action, Starting the Prompt Design Site: A New Home in our Stack Exchange Neighborhood. In fact Typically, whenwe want to actually compute this integral we have to write it as an iteratedintegral. Choose $\delta_k \ (k\in \{1,\ldots,n_0\})$ such that $|x-x_0|<\delta_k$ implies $|f_k(x)-f_k(x_0)|<\epsilon$. Suppose first that there exists a continuous function $\tilde{f}:[a, b] \rightarrow \mathbb{R}$ such that $f=\tilde{f}_{\mid(a, b)}$. Let $D$ be a nonempty subset of $\mathbb{R}$. Then $x,y \in (0,1)$ and $|x-y|=\delta_{0}<\delta$, but, \[|f(x)-f(y)|=\left|\frac{1}{x}-\frac{1}{y}\right|=\left|\frac{y-x}{x y}\right|=\left|\frac{\delta_{0}}{2 \delta_{0}^{2}}\right|=\left|\frac{1}{2 \delta_{0}}\right| \geq 2=\varepsilon .\]. Convolution - Wikipedia then $\lim_{x \to 0^+}f*f(x) = \infty$ and $\lim_{x \to 0^-}f*f(x)=0$. Are there any other agreed-upon definitions of "free will" within mainstream Christianity? Are Prophet's "uncertainty intervals" confidence intervals or prediction intervals? Stack Exchange network consists of 182 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. \nonumber\], Let $D$ be a nonempty subset of $\mathbb{R}$. When can it be possible if it is not possible with the given conditions? The following example illustrates this point. Is this an equivalent definition of uniform continuity? Let $\varepsilon \gt 0$. \end{array}\right .\], By its definition $\tilde{f}_{\mid(a, b)}=f$ and, so, $tilde{f}$ is continuous at every $x \in (a,b)$. If not, what are counter-examples? Choose $\delta > 0$ such that $|f(u)-f(v)|<\varepsilon$ whenever $u,v \in D$ and $|u-v|<\delta$. Edit: I accidentally left out the characteristic function on $j_{\epsilon}(x)$. \end{aligned} .\], Note that one can justify that inequality, \[\frac{\sqrt{|u|+|v|}}{\sqrt{u}+\sqrt{v}} \leq 1\]. 0. Assume that both $f(x)$ and $g(y)$ are defined for all real numbers. Clearlyfn; gn!xuniformly onR. Are there any MTG cards which test for first strike? The following theorem offers a sequential characterization of uniform continuity analogous to that in Theorem 3.3.3. Claim 1: the function x a(x) a(x + t) x a ( x) a ( x + t) is bounded (by M M) and uniformly continuous. Consider the function $a(s)=\dfrac{1}{1+s^2}$ and the space $X=\{f:\mathbb{R}\to \mathbb{R}$ such that $t\mapsto a(t)f(t)$ is bounded uniformly continuous$ \}$. Choose $n_0$ such that $|f_n(x)-f_{n_0}(x)|<\epsilon$ for all $x$ and all $n\ge n_0$. Does teleporting off of a mount count as "dismounting" the mount? (2) To prove this make the change of variablet=xyin the integral (1). Then $f$ is not Lipschitz continuous on $D$, but it is Hlder continuous on $D$ and, hence, $f$ is also uniformly continuous on this set. Statement from SO: June 5, 2023 Moderator Action, Starting the Prompt Design Site: A New Home in our Stack Exchange Neighborhood. I have a theorem that says if $g_i$ satisfy the conditions as above and $f$ be bounded and integrable over $\mathbb{R}^n$, then $g_i*f\to f$ over $\mathbb{R}^n$ with the convergence being uniform over all compact subsets of $\mathbb{R}^n$. Then for every $f\in L^p(\mathbb{R}^n)$, $1\le p <\infty$, $$\lim_{k\to\infty}||f*\phi_k-cf||_p=0$$. Then by the convolution theorem, $h\equiv f\star g$ has Fourier coefficients $\hat{h_n}=2\pi\hat{f_n}\hat{g_n}$. This shows $f$ is not uniformly continuous on $(0,1)$. Let $D$ be a nonempty subset of $\mathbb{R}$ and $f: D \rightarrow \mathbb{R}$. Moreover, $\lim _{x \rightarrow a^{+}} \tilde{f}(x)= \lim _{x \rightarrow a^{+}} f(x)=\tilde{f}(a)$ and $\lim _{x \rightarrow b^{-}} \tilde{f}(x)=\lim _{x \rightarrow b^{-}} f(x)=\tilde{f}(b)$, so $\tilde{f}$ is also continuous at $a$ and $b$ by Theorem 3.3.2. since $|f'_n(x)|$ is continuous on $[a,b]$. If a function $f: D \rightarrow \mathbb{R}$ is Hlder continuous, then it is uniformly continuous. A function $f: D \rightarrow \mathbb{R}$ is said to be Hlder continuous if there are constants $\ell \geq 0$ and $\alpha > 0$ such that, \[|f(u)-f(v)| \leq \ell|u-v|^{\alpha} \text { for every } u, v \in D .\]. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Are there any other agreed-upon definitions of "free will" within mainstream Christianity? Then the function $f(x)=\sqrt{x}$ is Lipschitz continuous on $D$ and, hence, uniformly continuous on this set. Then there exists $\varepsilon_{0}>0$ such that for any $\delta > 0$, there exists $u,v \in D$ with, Thus, for every $n \in \mathbb{N}$, there exists $u_{n}, v_{n} \in D$ with, Since $D$ is compact, there exist $u_{0} \in D$ and a subsequence $\left\{u_{n_{k}}\right\}$ of $\left\{u_{n}\right\}$ such that, \[u_{n_{k}} \rightarrow u_{0} \text { as } k \rightarrow \infty .\], \[\left|u_{n_{k}}-v_{n_{k}}\right| \leq \frac{1}{n_{k}} .\], \[v_{n_{k}} \rightarrow u_{0} \text { as } k \rightarrow \infty .\], \[f\left(u_{n_{k}}\right) \rightarrow f\left(u_{0}\right) \text { and } f\left(v_{n_{k}}\right) \rightarrow f\left(u_{0}\right) .\]. Another advantage of CNN is that it requires easy training. Does Pre-Print compromise anonymity for a later peer-review? because we know that there's a constant $M_t$ such that for all $s\in \mathbb{R} $ Now set = / 6. Write Query to get 'x' number of rows in SQL Server. mean value of an integral converges to function value. The choice of which function is reflected and shifted before the integral does not change the integral result (see commutativity ). $$as desired. I want to show that $X$ is translation invariant, i.e. $$ Show that the convolution $f\ast K$ is a uniformly continuous and bounded function. Prove that there exists $x_{0} \in[a, \infty)$ such that. How to skip a value in a \foreach in TikZ? \int_{|y|\geq\delta} |f(x-y)-f(x)||j_\epsilon(y)|dy\leq 2\sup_{x\in\mathbb{R}}|f(x)|\cdot\int_{|z|>\frac{\delta}{\epsilon}}|j(z)|dz. Keeping DNA sequence after changing FASTA header on command line. '90s space prison escape movie with freezing trap scene. by squaring both sides since they are both positive. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Does "with a view" mean "with a beautiful view"? Early binding, mutual recursion, closures. Now, if we make $f$ uniform continuous, how can we ensure uniform convergence throughout $\mathbb{R}^n . We observe first that, \[\begin{align*} \mid \frac{u^{2}}{u^{2}+1} -\frac{v^{2}}{v^{2}+1} \mid &= | \frac{u^{2}\left(v^{2}+1\right)-v^{2}\left(u^{2}+1\right)}{\left(u^{2}+1\right)\left(v^{2}+1\right)} | \\[4pt] &= \frac{|u-v||u+v|}{\left(u^{2}+1\right)\left(v^{2}+1\right)} \leq \frac{|u-v|(|u|+|v|)}{\left(u^{2}+1\right)\left(v^{2}+1\right)} \\[4pt] &\leq \frac{|u-v|\left(\left(u^{2}+1\right)+\left(v^{2}+1\right)\right)}{\left(u^{2}+1\right)\left(v^{2}+1\right)} \\[4pt] &\leq|u-v|\left(\frac{1}{v^{2}+1}+\frac{1}{u^{2}+1}\right) \leq 2|u-v| , \end{align*}\]. The constant $c$ is multiplied to make $\int j_\epsilon(x)dx = 1$. This page titled 3.5: Uniform Continuity is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Lafferriere, Lafferriere, and Nguyen (PDXOpen: Open Educational Resources) . $$ First, assume that $K$ is continuous with compact support. rev2023.6.27.43513. Problem involving number of ways of moving bead. Now set $\delta=\varepsilon / 2$. A function $f:(a, b) \rightarrow \mathbb{R}$ is uniformly continuous if and only if $f$ can be extended to a continuous function $\tilde{f}:[a, b] \rightarrow \mathbb{R}$ (that is, there is a continuous function $\tilde{f}:[a, b] \rightarrow \mathbb{R}$ such that $f=\tilde{f}_{\mid(a, b)}$). Royden, Real Analysis, Proposition 8 of Chapter 4, on p.128 of the third edition. We apply this result to stu Uniform equicontinuity, multiplier topology and continuity of convolution | SpringerLink $$, $$ Then we conclude by a density argument: if $K_n\to K$ in $L^1$, then $(f\star K_n)_n$ converges uniformly on the real line to $f\star K$. Then, whenever $u,v \in D$, with $|u-v|<\delta$ we have, \[|f(u)-f(v)| \leq \ell|u-v|^{\alpha}<\ell \delta^{\alpha}=\varepsilon . That might not be true even if $f_n$ converges uniformly to a function. $$ Let $a,b \in \mathbb{R}$ and $a < b$. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. rev2023.6.27.43513. By Arzela-Ascoli, it is equicontinous. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Then, if $u, v \in(a, b)$, $|u-a|<\delta$, and $|v-a|<\delta$ we have, \[|u-v| \leq|u-a|+|a-v|<\delta+\delta=\delta_{0}\], and, hence, $|f(u)-f(v)|<\varepsilon$. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. 0& \text{otherwise} Can I just convert everything in godot to C#. $$ \sum_{n\in\mathbb{Z}} |\hat{h_n}|=\sum_{n\in\mathbb{Z}} 2 \pi |\hat{f_n}\hat{g_n}|\le 2\pi\langle f,g\rangle <\infty$$ Keeping DNA sequence after changing FASTA header on command line. \lim_{\epsilon\to 0}\lVert(f*j_\epsilon)(\cdot)-f(\cdot)\rVert_\infty =0, the convolution of integrable functions is continuous? Convolution of probability distributions - Wikipedia Let > 0 > 0 and x0 x 0 be given. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. by Parseval's identity. Uniformly convergent implies equicontinuous, Statement from SO: June 5, 2023 Moderator Action, Starting the Prompt Design Site: A New Home in our Stack Exchange Neighborhood. It should just be: f ( x) = g ( x) = { x 3 / 4 x > 0 0 x 0. Learn more about Stack Overflow the company, and our products. Where in the Andean Road System was this picture taken? How to exactly find shift beween two functions. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. What is the best way to loan money to a family member until CD matures? fourier analysis - convolution a continuous function? - Mathematics I thought I used Holder's inequality correctly, since I was considering some fixed $x$, meaning that I can consider $f(x-y)$ as a function in $y$. What does the editor mean by 'removing unnecessary macros' in a math research paper? You don't need to multiply $\chi_{[-1,1]}(x)$, though it doesn't matter for your purpose. My idea is to use uniform convergence to deal with the "tail" and then use continuity to deal with the finitely many $f_n$'s left. We will show that $f$ is not uniformly continuous on $(0,1)$. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. How to exactly find shift beween two functions? rev2023.6.27.43513. Then $f$ is uniformly continuous on $D$. Does teleporting off of a mount count as "dismounting" the mount? We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. I see this question is a special case with $p=\infty$, but unfortunately the theorem states that $1\le p <\infty$. Does Pre-Print compromise anonymity for a later peer-review? The following result is straightforward from the definition. Therefore, $f$ converges to zero, which is a contradiction. Suppose f L ( R) and K L 1 ( R) with R K ( x) d x = 1. One natural operation is multiplication of functions, but unfortunately L1(R) is not closed under multiplication. Dene theconvolution (f g)(x):=Z(1)f(xy)g(y)dy One preliminary useful observation is g=gf. convolution of compactly supported continuous function with schwartz class function is again a Schwartz class function? This follows directly from, Show that convolution of two $L^1(\mathbb{R})$ functions is continuous, Statement from SO: June 5, 2023 Moderator Action, Starting the Prompt Design Site: A New Home in our Stack Exchange Neighborhood. It follows that $\{\hat{h_n}\}$ is absolutely summable, since. |(f \ast g)(x_{n}) - (f \ast g)(x)| \leq \Vert f(\cdot - x) - f(\cdot - x_{n})\Vert_{\infty} \Vert g \Vert_{1} \lt \varepsilon \cdot \Vert g \Vert_1 Is there a lack of precision in the general form of writing an ellipse? 1.3 Convolution 15 1.3 Convolution Since L1(R) is a Banach space, we know that it has many useful properties.In particular the operations of addition and scalar multiplication are continuous. If one density function is Gaussian and the other is uniform, their convolution is a 'blurred gaussian'. Thank you! Is there an extra virgin olive brand produced in Spain, called "Clorlina"? Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. Since the difference between $f*g(x-a)$ and $f*g(x)$ for any fixed $x$ and any $a$ is bounded, I can choose $a$ such that $||h||_q < \frac{\varepsilon}{|| f ||_p}$, so $|f*g(u) - f*g(x)| < \varepsilon$ for any arbitrary $\varepsilon > 0$ and any $u \in (x-a, x+a)$. $f(x)=1 / x \text { on }[a, \infty)$, where $a > 0$. rev2023.6.27.43513. rev2023.6.27.43513. Now define, $\tilde{f}:[a, b] \rightarrow \mathbb{R}$ by, \[\tilde{f}(x)=\left\{\begin{array}{ll} Learn more about Stack Overflow the company, and our products. While every uniformly continuous function on a set $D$ is also continuous at each point of $D$, the converse is not true in general. If $f\in L^{\infty}(\mathbb{R}^n)$ and $f$ is continuous at $x$, then $$\lim_{k\to\infty}(f*\phi_k)(x)=cf(x)$$ If $f\in L^{\infty}(\mathbb{R}^n)$ and is uniformly continuous, then $f*\phi_k\to cf$ uniformly; that is, $$\lim_{k\to\infty}||f*\phi_k-cf||_{\infty}=0$$. But what about uniform continuity? What are these planes and what are they doing? If not, what are counter-examples? The condition is named after Otto Hlder . Can you legally have an (unloaded) black powder revolver in your carry-on luggage? \lim_{x\to y} h(x) &= \lim_{x\to y} \sum_{n\in\mathbb{Z}} \hat{h_n}e^{inx} \\ The best answers are voted up and rise to the top, Not the answer you're looking for? Convolution of a sequence of L1 function has uniformly convergence. If = 1, then the function satisfies a Lipschitz condition. real analysis - Uniform continuity and translation invariance Otherwise you'll need that the supports of the $g_i$ "tend" to $\{0\}$ and $A=\sup_i \|g_i\|_{L^1}<\infty$. 3.7: Transformations of Random Variables - Statistics LibreTexts If $f\in L^{\infty}(\mathbb{R}^n)$ and $f$ is continuous at $x$, then $$\lim_{k\to\infty}(f*\phi_k)(x)=cf(x)$$ If $f\in L^{\infty}(\mathbb{R}^n)$ and is uniformly continuous, then $f*\phi_k\to cf$ uniformly; that is, $$\lim_{k\to\infty}||f*\phi_k-cf||_{\infty}=0$$ Is a naval blockade considered a de-jure or a de-facto declaration of war? To subscribe to this RSS feed, copy and paste this URL into your RSS reader. First observe that for u, v [ 3, 2] we have | u + v | | u | + | v | 6. Let $f: D \rightarrow \mathbb{R}$ be a continuous function. The architecture of CNN is developed to yield benefit of two- dimensional input image. I have a theorem that says if $g_i$ satisfy the conditions as above and $f$ be bounded and integrable over $\mathbb{R}^n$, then $g_i*f\to f$ over $\mathbb{R}^n$ with the convergence being uniform over all compact subsets of $\mathbb{R}^n$. Share Cite Improve this answer For $n$ so large that $|x_{n} - x| \lt \delta$ we then have Would limited super-speed be useful in fencing? I know the convolution of a $L^1$ and a $L^{\infty}$ function is uniformly continuous. $$ There is the inequality $\|f\ast K\|_\infty\leq\|f\|_\infty\|K\|_1$, which yields that $f\ast K$ is bounded. Problem involving number of ways of moving bead. The best answers are voted up and rise to the top, Not the answer you're looking for? Apr 23, 2022 3.6: Distribution and Quantile Functions 3.8: Convergence in Distribution Kyle Siegrist University of Alabama in Huntsville via Random Services Table of contents Basic Theory The Problem Transformed Variables with Discrete Distributions Transformed Variables with Continuous Distributions The Change of Variables Formula \end{eqnarray}$$, $$ Alternative to 'stuff' in "with regard to administrative or financial _______.". $\endgroup$ But I have problem bounding the residual, which would be of form: $$ \int_\mathbb{R} |f(x-z) - f_M(x-z)|\cdot |g(z)| dz$$. 3 Answers. Combining every 3 lines together starting on the second line, and removing first column from second and third line being combined. Proving that the smooth, compactly supported functions are dense in $L^2$. We are going to prove that, \[|f(u)-f(v)| \leq|u-v|^{1 / 2} \text { for every } u, v \in D .\], The inequality in (3.9) holds obviously for $u = v = 0$. For $u > 0$ or $v > 0$, we have, \[\begin{aligned} It is defined as the integral of the product of the two functions after one is reflected about the y-axis and shifted. Would you care explaining what's the downvote for? \[ The use of Hlder's inequality is fine since $\|f(x-y)\|_p = \|f(y)\|_p$ by translation invariance of the integral. How can I know if a seat reservation on ICE would be useful? Or is having a compact support not necessary here? &=\left|\frac{u-v}{\sqrt{u}+\sqrt{v}}\right| \\ PDF convolution - University of Pennsylvania Can you legally have an (unloaded) black powder revolver in your carry-on luggage? Then using Finally, by taking $\delta \to 0$, we get if $f\in X$ then $f_t\in X$ for all $t\in \mathbb{R}$, where $f_t$ is the $t$-translation of $f$ defined by $f_t(s)=f(t+s)$. As $\epsilon \to 0$, we have $\frac{\delta}{\epsilon}\to \infty$, and hence that Two other questions: how is the variable $z$ defined, and how is the limsup inequality justified? But I'm having trouble writing it down. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. &=\sqrt{|u-v|} \frac{\sqrt{|u-v|}}{\sqrt{u}+\sqrt{v}} \\ Theorem 11.5 (Continuous Functions are Dense). Thanks beforehand. \int_{|y|<\delta} |f(x-y)-f(x)||j_\epsilon(y)|dy\leq \lVert j\rVert_1\cdot\sup_{x\in\mathbb{R},|y|<\delta}|f(x)-f(x-y)|.