Can you have a constant dependent on epsilon when bounding expressions?

Question

Not sure how to best phrase the question, so I'll simply use a problem I solved that slightly confused me. I need to prove the following claim:

Claim. Let $1 \le p < \infty$ and $f \in L^p(\mathbb{R}, \mathcal{L}, \mu)$ where $\mu$ is the Lebesgue measure. For $h>0$ define $f_h(x) = \frac{1}{h} \int_{x}^{x+h}f(t)\ dt$. Prove that $\lVert f_h - f \rVert_p \to 0$ as $h \to 0$.

Proof. Let $\varepsilon > 0$. Since $C_K(\mathbb{R})$ (continuous functions with compact support) is dense in $L^p$, let $g \in C_K(\mathbb{R})$ such that $\|f - g\|_p < \frac{\varepsilon}{3}$ and $g = 0$ on $\mathbb{R} \setminus K$ where $K$ is compact. We have \begin{equation} \|f_h - f\|_p \leq \|f_h - g_h\|_p + \|g_h - g\|_p + \|g - f\|_p \end{equation} where $g_h(x) = \frac{1}{h}\int_x^{x+h} g(t) \, dt$.

Observe that $f_h(x) - g_h(x) = \int_x^{x+h} (f(t) - g(t)) \, dt = (f - g)_h(x)$. Since $\| f_h \|_p \le \| f \|_p$ (I have proven this in a previous problem), thus $\|f_h - g_h\|_p \leq \|f - g\|_p < \frac{\varepsilon}{3}$.

Let $K \subset [a, b]$ for some $a < b$. Note that $g$ is uniformly continuous on $[a-1, b+1]$, so suppose \begin{equation} |g(x) - g(t)| < \frac{\varepsilon}{3 (b-a+1)^{1/p}} \end{equation} whenever $|x - t| < \delta$ for all $x, t \in [a-1, b+1]$. Accordingly, let $h < \min\{\delta, 1\}$.

Since $g = 0$ on $\mathbb{R} \setminus [a, b]$, observe that $h < 1$ implies $g_h$ is $0$ on $\mathbb{R} \setminus [a-1, b]$. Thus, \begin{align*} \|g_h - g\|_p^p &= \int_{a-1}^{b} \left| g_h(x) - g(x) \right|^p \, dx \\ &= \int_{a-1}^{b} \left| \frac{1}{h} \int_x^{x+h} (g(t) - g(x)) \, dt \right|^p dx \\ &\leq \int_{a-1}^{b} \frac{1}{h^p} \left( \int_x^{x+h} |g(t) - g(x)| \, dt \right)^p dx \\ &\leq \int_{a-1}^{b} \frac{1}{h^p} \left( \int_x^{x+h} \frac{\varepsilon}{3 (b-a+1)^{1/p}} \, dt \right)^p dx \\ &= \int_{a-1}^{b} \frac{1}{h^p} \left( \frac{\varepsilon}{3 (b-a+1)^{1/p}} \cdot h \right)^p dx \\ &= \int_{a-1}^{b} \frac{1}{h^p} \cdot \frac{\varepsilon^p}{3^p (b-a+1)} \cdot h^p dx \\ &= \frac{\varepsilon^p}{3^p (b-a+1)} \cdot \int_{a-1}^{b} \, dx \\ &= \frac{\varepsilon^p}{3^p}. \end{align*}

Taking the $p$-th root, we get $\|g_h - g\|_p \le \frac{\epsilon}{3}$ which completes the proof. $\Box$

Now, in initially doing this proof, I originally did:

Suppose $|g(x) - g(t)| < \epsilon$ whenever $|x - t| < \delta$ for all $x, t \in [a-1, b+1]$. This made it so my bound on $\|g_h - g \|_p$ and thus $\|f_h - f \|_p$ depends on $(b-a+1)$ which depends on $\epsilon$ (since $g$ and thus $K$ themselves depend on the $\epsilon$). Just looking at it as: $\|f_h - f \|_p < \epsilon(b-a+1)$ makes me think it's wrong since the $b-a+1$ depends on $\epsilon$, so we do not know if the whole expression goes to $0$ as $\epsilon \to 0$, but if we adjust the uniform continuity bound, this term cancels.

This doesn't feel completely intuitive to me and almost like I'm missing something, but I'm not even sure how to best ask this. It's almost like, I am not fully sure when bounding things if it's allowed for constants to be dependent on $\epsilon$, it seems more nuanced than this, and I would like to have a better approach of what exactly can and cannot depend on $\epsilon$. My current approach is if I can fix the bounds such that everything in the end simplifies nicely to $\epsilon$ then clearly we are fine, but as problems get more complex, it would be nice to be able to have more messy $\epsilon$ bounds containing constants and be able to just say: letting $\epsilon \to 0$ we achieve what we want.

Edit. Compare this to the following: We want to show $\lim_{x \to 0} |f(x)| = 0$. Let $\epsilon > 0$. Let $N(\epsilon)$ be some constant dependent on $\epsilon$. Now, suppose we know for some $\delta(N, \epsilon)$ we have $|f(x)| < N\epsilon$ whenever $|x| < \delta$. In this case it's clear that this is not enough since $\epsilon N(\epsilon)$ may not go to $0$. What makes this different than the prior $L^p$ problem where even if we have the bound depending on some constant($\epsilon$) it so happened to work out nicely in that case? Is there a better way of thinking of the original $L^p$ problem that makes it clear why having the expression $< (b-a+1)$ would still be fine?

Answer 1

To long for a comment. Not sure this answers your question.

In your proof you make the following choices, where later choices depend on earlier ones: $\epsilon \leadsto g \leadsto (a,b) \leadsto \delta \leadsto h$.

In the correct version of the proof, you choose $\delta$ such that $(b-a)$ cancels at the end.

In the 'initial' version of the proof, you end up with an estimate $something \le \epsilon (b-a)$. This would not imply the claim as we cannot show that $b-a$ stays bounded for $\epsilon\to0$. Then you were right to backtrack in the proof and to adjust the choice of $\delta$.

To me it seems, that you did not missed anything. Instead you realized that the initial proof was not correct and you realized how to fix it. (Realizing is the opposite of missing!)

Books, lecture notes, papers etc often present finished proofs, where all the constants magically fit together. This shows the final result not the evolution. The evolution of the proof for this $L^p$-result would be pretty much what you described.

Answer 2

The estimates you wrote in the main body cannot really be simplified, because of the manner in which you structured it. But see below for another presentation which avoids such issues.

Having an $N(\epsilon)$ in your edit is also bad because we cannot a-priori guarantee that $N(\epsilon)\epsilon\to 0$ as $\epsilon\to 0^+$. The following may be elementary, but I suggest reviewing it. Here I go over twelve or so equivalent statements of limits. Furthermore, you’re right that always phrasing things exactly in terms of $\delta$’s can get a little annoying. Once you get the hang of things, we can finesse some proofs using $\limsup$ or $\liminf$ to make proofs shorter. The ‘meta’ reason is that $\limsup$ (and $\liminf$) always exist, whereas $\lim$ does not always. Just in case, here is a quick review of the very general definition of $\limsup$, but really the rest of my answer assumes you know how to work with $\limsup$.

Before we get into some more ‘advanced’ stuff (but really it all becomes obvious very quickly) let’s note that as mentioned in the comments, whether we use $\epsilon$ in the intermediate step of some multiple of it, really depends on a case-by-case basis.

• The way you wrote your proof, you cannot use $\epsilon$ alone when bounding $|g(x)-g(t)|$.
• In a measure-theory proof, for say proving countable union of measure-zero sets have measure-zero, we cannot avoid (some variant of) the “$\frac{\epsilon}{2^n}$ trick”, because we have to add up infinitely many terms.
• When dealing with uniform convergence, (say to prove that uniform limits of continuous functions are continuous), one typically employs the “$\frac{\epsilon}{3}$ trick”. However, here we can simply fix $\epsilon$ at each stage and end up with $3\epsilon$ at the end, and that’s perfectly fine.

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Writing $\epsilon$-$\delta$ proofs can get annoying, and indeed very troublesome. I can think of a few ways of shortening them, by combining some or all of the following:

• break up the problem into simpler steps, then put it back together. Particularly important when dealing with estimates arising from some dense subcollection of functions.
• use $\limsup$ and $\liminf$ whenever possible, because they always exist. This way you can avoid the struggle of “how exactly I choose my $\widetilde{\epsilon}$ in terms of $\epsilon$”. The trade-off here is that often we’ll have a bound depending on several variables, and we must decide the correct order in which to apply the $\limsup$’s to kill of various terms (see the Cesaro example below).
• Since it’s so important, I’ll mention again the importance of density!

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Example 1. Your $L^p$ example.

You’re trying to prove that $f\in L^p$ implies $\|f_h-f\|_p\to 0$ as $h\to 0^+$, but this is pretty hard a-priori. So, instead, break it up into several steps:

• Invoke the general fact that $C_c(\Bbb{R})$ is dense in $L^p(\Bbb{R})$ for $1\leq p<\infty$.
• Prove that if $f\in L^p$ then $\|f_h\|_p\leq \|f\|_p$ (actually it would be fine even if we had a bound like $\|f_h\|_p\leq \psi_0(\|f\|_p)$ where $\psi_0(x)$ is a function which vanishes as $x\to 0^+$).
• prove that if $g\in C_c(\Bbb{R})$ then, we have $\|g_h-g\|_p\to 0$. The benefit of doing this separately is that we’re now fixing the function $g$ first of all and only afterwards do we introduce $\epsilon$. So, now if you have an estimate like:

there is a ‘big’ constant $B_g$ (like its sup norm, size of support etc) and a function $\psi_1(\epsilon)$ with vanishing limit at the origin (e.g $\epsilon^p$ or $\frac{\epsilon^p}{p}$ or what have you), such that for all (sufficiently small) $\epsilon>0$ there is a $\delta>0$ such that for all $0<h<\delta$, we have $\|g_h-g\|_p\leq B_{g}\psi_1(\epsilon)$,

then this falls into one of the twelve equivalences above, and so we can conclude that $\|g_h-g\|_p\to 0$ as $h\to 0^+$.

With these three facts, we can deduce the general claim because for any $f\in L^p$, any $g\in C_c$, any $\epsilon>0$, and any $h>0$, we have \begin{align} \|f_h-f\|_p&\leq \|(f-g)_h\|_p+\|g_h-g\|_p+\|g-f\|_p\\ &\leq \psi_0(\|f-g\|_p)+ \|g_h-g\|_p+\|g-f\|_p.\tag{$*$} \end{align} First, take the $\limsup\limits_{h\to 0^+}$ of both sides, and use the third bullet point to deduce that the middle term vanishes: \begin{align} \limsup_{h\to 0^+}\|f_h-f\|_p&\leq \psi_0(\|f-g\|_p)+\|g-f\|_p. \end{align} Now, by density of $C_c$ in $L^p$ and by arbitrariness of $g$, the infimum of the RHS over all $g\in C_c$ is $0$ (if you want, take a sequence of $g$’s in $C_c$ which converge to $f$ in $L^p$, then say). This shows \begin{align} \limsup_{h\to 0^+}\|f_h-f\|_p&\leq 0, \end{align} and thus it equals $0$. Since the $\limsup$ of a non-negative function is $0$, the limit exists and is also $0$. With some practice, it becomes immediate to observe from $(*)$ and the third bullet point that $\|f_h-f\|_p\to 0$ as $h\to 0^+$.

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Example 2: Fourier-ish stuff and Density

Look at Show that $\lim_{N\to\infty}\int_0^1 \left|\frac1N\sum_{n=1}^N f(x+\xi_n) \right|^{\,2}\, dx = 0$ if $\int_0^1 f(x)\, dx = 0$ and $f$ is periodic. The intermediate details aren’t really important, the key I want you to focus on is:

• the definition of $T_N$ as some operator between some function spaces in the beginning (similar to your $f\mapsto f_h$)
• $T_N$ is linear and uniformly bounded; $\|T_Nf\|_2\leq \sqrt{2}\|f\|_2$, similar to your $\|f_h\|_p\leq \|f\|_p$.
• we exploited density by choosing a collection of functions where we know that for all $g\in\mathcal{G}$, we have $\|T_Ng\|_2\to 0$ as $N\to\infty$ (similar to your $\|g_h-g\|_p\to 0$ for $g\in C_c$).

Writing up all of these together in a single $\epsilon$-$N$ proof would be an absolute nightmare.

Here is a similar story when trying to prove the Riemann-Lebesgue lemma.

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Example 3: Cesaro-type limits.

Theorem.

Let $f\in L^1_{\mathrm{loc}}(\Bbb{R})$ and suppose $\lim\limits_{x\to\infty}f(x)=\ell\in\Bbb{R}$. Then, $\lim\limits_{x\to\infty}\frac{1}{x}\int_0^xf(t)\,dt=\ell$.

The proof is pretty easy with the right framing. Let $\epsilon>0$; then by hypothesis there is an $R>0$ such that for all $x\geq R$ we have $|f(x)-\ell|\leq \epsilon$. Now, for all $x\geq R$, we have \begin{align} \left|\frac{1}{x}\int_0^xf(t)\,dt-\ell\right|&\leq\left| \frac{1}{x}\int_0^Rf(t)\,dt\right|+\left|\frac{1}{x}\int_R^x[f(t)-\ell]\,dt\right|+\left|\frac{1}{x}\int_R^x\ell\,dt-\ell\right|\\ &\leq \frac{1}{x}\left|\int_0^Rf(t)\,dt\right|+\frac{\epsilon|x-R|}{x}+ \frac{|M\ell|}{x} \tag{$**$}. \end{align} Trying to get a full $\epsilon$-$N$ proof from here would be pretty annoying because the $N$ would have to be chosen correctly in terms of $R$ and $\epsilon$ and so on (see the top voted answer here). However, now we can simply take the $\limsup\limits_{x\to\infty}$ of both sides to deduce that the first and third terms vanish while the second is $\epsilon$: \begin{align} \limsup_{x\to\infty}\left|\frac{1}{x}\int_0^xf(t)\,dt-\ell\right|&\leq\epsilon. \end{align} Finally, since $\epsilon>0$ was arbitrary, we conclude the LHS is actually $0$, and so once again, the $\limsup$ of a non-negative function being $0$ implies the limit exists and is $0$.

This sort of thing is how we would like to reason intuitively, and working with $\limsup$’s is the right way to go (this answer tries to do a similar thing, but strictly speaking it’s not right, because $\lim$ is introduced too early before even proving the limit exists). So once again, I’ll emphasize that once you get to the bound $(**)$, an “experienced analyst” will immediately conclude that as $x\to\infty$, the limit of the LHS exists and is $0$.

Btw, as an aside: note that we cannot take $\limsup\limits_{\epsilon\to 0^+}$ of both sides of $(**)$ first and then $\limsup\limits_{x\to\infty}$, because the $R$ depends on $\epsilon$, and so the domain of $x$’s also depends on $\epsilon$. This is what I meant earlier when I said that we trade in explicit intermediate bounds in terms of $\epsilon$ for a correct order of taking $\limsup$’s.

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There are several other examples of how in analysis, one can write arguments in such a way as to limit the amount of “explicit” crazy $\epsilon$-dependent estimates within substeps. But hopefully the above gives you an idea of some of the typical arguments one often uses in analysis (and which most advanced books on real/Fourier/harmonic analysis skip over).