# On the interchange of limit and sup (with also an integral)

Let $\{f_n\}$ be a sequence of continuous functions $f_n:\mathbb{R} \to \mathbb{R}$.

Suppose $\Vert f_n\Vert_\infty \le C$, $f_n \to f$ uniformly (we can also assume $f_n$ Lipschitz if necessary). Then

$$\lim_{n \to \infty} \sup_{1 +\epsilon < k \le \frac{3}{2}}\int_{\mathbb{R}}\frac{f_n(x+t) - f_n(x)}{|t|^{k}}dt = \sup_{1 +\epsilon < k \le \frac{3}{2}}\lim_{n \to \infty}\int_{\mathbb{R}}\frac{f_n(x+t) - f_n(x)}{|t|^{k}}dt.$$

• Is this claim true? How can I prove it? If it is not true, what assumptions should we add so that the claim holds?
• You've got some superfluous notation. What is $\epsilon$? – Umberto P. Apr 14 '17 at 19:40
• @UmbertoP. $\epsilon >0$ is fixed and small. – user414592 Apr 14 '17 at 19:42
• Sorry - small font and old eyes. The $3/2$ looked like $s/2$. – Umberto P. Apr 14 '17 at 19:43
• $x$ is fixed too? – Umberto P. Apr 14 '17 at 19:43
• @UmbertoP. No worries. – user414592 Apr 14 '17 at 19:43

Your claim is correct, if you add uniformly bounded Lipschitz constants to the list, as Gio67 hinted (see the remark below for a generalisiation).

Proof: First, I show that not only the integrals exist, but also their supremum on the left side of your equation. Let $L_n$ be the Lipschitz constant for $f_n$ with $L_n\leq L$ for each $n\in\mathbb{N}$. For $1+\epsilon<k\leq\frac3 2$ we have $$\int_\mathbb{R} \left|\frac{f_n(x+t)-f_n(x)}{|t|^k}\right|\,dt=\int_{[-1,1]} \frac{|f_n(x+t)-f_n(x)|}{|t|^k}\,dt+\int_{\mathbb{R}\setminus [-1,1]} \frac{|f_n(x+t)-f_n(x)|}{|t|^k}\,dt\leq\int_{[-1,1]} \frac{L_n\cdot|(x+t)-x|}{|t|^\frac3 2}\,dt+\int_{\mathbb{R}\setminus [-1,1]} \frac{2C}{|t|^{1+\epsilon}}\,dt=L_n\int_{[-1,1]}|t|^{-\frac1 2}\,dt+\int_{\mathbb{R}\setminus [-1,1]} \frac{2C}{|t|^{1+\epsilon}}\,dt.$$ This is equal to $$2L_n\int_{[0,1]}t^{-\frac1 2}\,dt+4C\int_{(1,+\infty)} |t|^{-(1+\epsilon)}\,dt=4L_n+\frac{4C}{\epsilon}\leq4L+\frac{4C}{\epsilon}.$$

Therefore all the integrals exist and $$\sup_{1+\epsilon<k\leq\frac3 2}\int_\mathbb{R} \frac{f_n(x+t)-f_n(x)}{|t|^k}\,dt\leq\sup_{1+\epsilon<k\leq\frac3 2}\int_\mathbb{R} \left|\frac{f_n(x+t)-f_n(x)}{|t|^k}\right|\,dt\leq4L+\frac{4C}{\epsilon}<+\infty.$$

Now, I claim that $$\lim_{n\to\infty}\int_\mathbb{R} \frac{f_n(x+t)-f_n(x)}{|t|^k}\,dt=\int_\mathbb{R} \frac{f(x+t)-f(x)}{|t|^k}\,dt,$$ uniformly in $k$. One has $$\int_\mathbb{R} \left|\frac{f_n(x+t)-f_n(x)}{|t|^k}-\frac{f(x+t)-f(x)}{|t|^k}\right|\,dt=\int_{[-1,1]} \frac{|(f_n(x+t)-f_n(x))-(f(x+t)-f(x))|}{|t|^k}\,dt+\int_{\mathbb{R}\setminus [-1,1]} \frac{|(f_n(x+t)-f_n(x))-(f(x+t)-f(x))|}{|t|^k}\,dt\leq\int_{[-1,1]} \frac{|(f_n(x+t)-f_n(x))-(f(x+t)-f(x))|}{|t|^{\frac3 2}}\,dt+\int_{\mathbb{R}\setminus [-1,1]} \frac{|(f_n(x+t)-f_n(x))-(f(x+t)-f(x))|}{|t|^{1+\epsilon}}\,dt;$$ since the last term is independent of $k$, it is sufficient to show that this last term converges to $0$ as $n\to\infty$. For the first one, note that from $|f_n(x)-f_n(y)|\leq L_n|x-y|\leq L|x-y|$ for each $n\in\mathbb{N}$ it follows that $|f(x)-f(y)|\leq L|x-y|$, hence that $f$ is also Lipschitz with constant $L$. Thus $$\frac{|(f_n(x+t)-f_n(x))-(f(x+t)-f(x))|}{|t|^{\frac3 2}}\leq\frac{2L|t|}{|t|^{\frac3 2}}=2L|t|^{-\frac1 2},$$ which is integrable on $[-1,1]$. Therefore $$\int_{[-1,1]} \frac{|(f_n(x+t)-f_n(x))-(f(x+t)-f(x))|}{|t|^{\frac3 2}}\,dt\to 0$$ by the dominated convergence theorem (the integrand converges to $0$ pointwise for $t\neq 0$, that is almost everywhere). For the second integral note that $f(x)\leq C$ and therefore $|(f_n(x+t)-f_n(x))-(f(x+t)-f(x))|\leq 4C$. It follows that $$\frac{|(f_n(x+t)-f_n(x))-(f(x+t)-f(x))|}{|t|^{1+\epsilon}}\,dt\leq\frac{4C}{|t|^{1+\epsilon}},$$ which is integrable on $\mathbb{R}\setminus [-1,1]$. Therefore, the integral converges to $0$, again by the dominated convergence theorem.

We hence are in the situation that functions $F_n:E\left(:=(1+\epsilon,\frac3 2]\right)\to\mathbb{R}$, $n\in\mathbb{N}$ converge uniformly to a function $F:E\to\mathbb{R}$ such that $$\sup_{k\in E}F(k)\in\mathbb{R}.$$ The goal is to conclude that $$\lim_{n\to\infty}\sup_{k\in E}F_n(k)=\sup_{k\in E}\lim_{n\to\infty}F_n(k),$$ especially that the limit on the left side exists.

For each $k_0\in E$ we evidently have $F_n(k_0)\leq\sup_{k\in E}F_n(k)$, so that $$\lim_{n\to\infty}F_n(k_0)\leq\liminf_{n\to\infty}\sup_{k\in E} F_n(k).$$ Hence ($k_0$ was arbitrary) $$\sup_{k\in E}\lim_{n\to\infty}F_n(k)\leq\liminf_{n\to\infty}\sup_{k\in E} F_n(k).\qquad (1)$$

On the other hand, let $\delta>0$ be arbitrary. Then there exists an integer $N\in\mathbb{N}$ so that $|F_n(k)-F(k)|\leq\delta$ whenever $k\in E$. For $m\geq N$, there exists a $k(m)\in E$ with $$\sup_{k\in E}F_m(k)-\delta\leq F_m(k(m))\leq F(k(m))+\delta\leq\sup_{k\in E}\lim_{n\to\infty}F_n(k)+\delta.$$ We conclude $$\sup_{k\in E}F_m(k)\leq\sup_{k\in E}\lim_{n\to\infty}F_n(k)+2\delta$$ and even ($\delta$ was arbitrary) $$\sup_{k\in E}F_m(k)\leq\sup_{k\in E}\lim_{n\to\infty}F_n(k).$$ We obtain $$\limsup_{n\to\infty}\sup_{k\in E}F_n(k)\leq\sup_{k\in E}\lim_{n\to\infty}F_n(k).\qquad (2)$$

(1) and (2) combined show that $$\liminf_{n\to\infty}\sup_{k\in E}F_n(k)=\limsup_{n\to\infty}\sup_{k\in E}F_n(k)=\sup_{k\in E}\lim_{n\to\infty}F_n(k),$$ in other words: $$\lim_{n\to\infty}\sup_{k\in E}F_n(k)=\sup_{k\in E}\lim_{n\to\infty}F_n(k),$$ the desired conclusion.

Remark: As the proof shows, we only need pointwise convergence and that the restrictions of the $f_n$ to $[-1,1]$ are Lipschitz with uniformly bounded Lipschitz constants. For example, locally Lipschitz (the restrictions to compact sets are Lipschitz) with uniformly bounded constants, at least for $[-1,1]$, is sufficient.

Addendum: the OP was interested in the following variant of the problem: is it correct that $$\lim_{n\to\infty}\sup_{1+\epsilon<k\leq\frac3 2}\int_\mathbb{R} \frac{f_n(x_n+t)-f_n(x_n)}{|t|^k}\,dt=\sup_{1+\epsilon<k\leq\frac3 2}\lim_{n\to\infty}\int_\mathbb{R} \frac{f_n(x_n+t)-f_n(x_n)}{|t|^k}\,dt,$$ if $f_n, f$ are as above and $x_n\to x$? The answer is yes, by a similar proof to above. By the lemma, it suffices to show that $$\int_\mathbb{R} \frac{f_n(x_n+t)-f_n(x_n)}{|t|^k}\,dt\to\int_\mathbb{R} \frac{f(x+t)-f(x)}{|t|^k}\,dt,$$ uniformly in $k$. In complete analogy to above, I can split the integral $$\int_\mathbb{R} \left|\frac{f_n(x_n+t)-f_n(x_n)}{|t|^k}-\frac{f(x+t)-f(x)}{|t|^k}\right|\,dt$$ into the integrals over $[-1,1]$ and $\mathbb{R}\setminus [-1,1]$ and estimate them by the corresponding integrals with $k=\frac3 2$ and $k=1+\epsilon$, respectively. If I could show that those intgrals converge to $0$, I would be done, just as above. Again in complete analogy to my answer, I can apply LDT to reach this conclusion if the integrands converge pointwise, i.e. $$f_n(x_n+t)\to f(x+t)\quad \text{and}\quad f_n(x_n)\to f(x),$$ for fixed $t$. Clearly (setting $t=0$), I only have to prove the first assertion. Since $(x_n)$ converges, it is a bounded sequence so that $K:=\overline{\{x_n:n\in\mathbb{N}\}}$ is compact, as well as $K+t$. $f_n$ converges locally uniformly, hence uniformly on $K+t$. We get $$|f_n(x_n+t)-f(x+t)|\leq |f_n(x_n+t)-f(x_n+t)|+|f(x_n+t)-f(x+t)|\leq ||f_n-f||_{\infty, K+t}+|f(x_n+t)-f(x+t)|\to 0,$$ since $x_n\to x$ and by the continuity of $f$. Now I have everything I need, as outlined above.

• Thanks for your answer. Does the result hold if $f_n \to f$ only locally uniformly (that is: uniformly on compact sets)? – user414592 May 11 '17 at 19:03
• @Ryo If you assume that the Lipschitz constants are uniformly bounded, the answer is most definitely yes, my proof shows that pointwise convergence is in fact sufficient. If you want to replace this condition, I strongly doubt it, I can't offer an example at the moment though...in this case, the supremum of the integrals on the left hand side could in general be infinity, but the one on the right hand side might be as well... – haemi May 12 '17 at 19:41
• Can you also show that $$\lim_{n \to \infty} \sup_{1 +\epsilon < k \le \frac{3}{2}}\int_{\mathbb{R}}\frac{f_n(x_n+t) - f_n(x_n)}{|t|^{k}}dt = \sup_{1 +\epsilon < k \le \frac{3}{2}}\int_{\mathbb{R}}\frac{f(x+t) - f(x)}{|t|^{k}}dt$$ if $f_n \to f$ locally uniformly and $x_n \to x$? – user414592 May 15 '17 at 22:14
• @Ryo Since $x_n$ is convergent, hence bounded, $\{x_n:n\in\mathbb{N}\}$ is included in a compact set so that $f_n(x_n)$ and $f_n(x_n+t)$ converge to $f(x)$ and $f(x+t)$, respectively, because $f_n$ is locally uniformly convergent. Now, in analogy to my answer, we can split up the integrals of the absolute value of the difference of the quotients in question, estimate them by the corresponding integrals with $k=\frac{3}{2}$, resp. $k=1+\epsilon$ and apply LDT to get convergence to zero. In other words, we have uniform convergence in $k$, so an application of my "lemma" yields the result. – haemi May 17 '17 at 20:11
• Thanks! Could you write the details of your reasoning and add this as an appendix to your answer? – user414592 May 17 '17 at 23:22

The integral is not well defined. Take $f_n(t)=f(t)=|t|^a$ with $a>0$ small. When $x=0$ you get $\frac{f(0+t)-f(0)}{|t|^k}=|t|^{a-k}$ which is not integrable near $t=0$. To make things work you probably need to assume that each function $f_n$ is Lispchitz. If you assume that the Lipschitz constants are uniformly bounded, then you can probably apply LDT.

• We can assume $f_n$ Lipschitz if necessary. My issue here is mostly the interchange of $\lim$ and $\sup$. – user414592 Apr 23 '17 at 12:14