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Suppose $(f_n)_{n >0}$ is a sequence of bounded continuous functions that converges locally uniformly to $f$. Is the following statement true?
$$\lim_n \sup_{\gamma \in (0,2)}\int_0^\infty \frac{f_n(x+t)-f_n(x)}{t^\gamma}dt = \sup_{\gamma \in (0,2)}\int_0^\infty \frac{f(x+t)-f(x)}{t^\gamma}dt $$

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  • $\begingroup$ Where are these functions defined, and is this supposed to be true for every x? $\endgroup$ – zhw. Mar 31 '17 at 20:54
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The statement is not true.

Consider the bounded and continuous functions $$f_n(x)= \frac{x-n}n1_{[n,n+1)}(x)+\frac{1}{n}1_{[n+1,\infty)}(x).$$ Then, for $x<1$,

\begin{eqnarray*}\lim_n\sup_{\gamma\in(0,2)}\int_0^\infty\frac{f_n(x+t)-f_n(x)}{t^\gamma}dt&\geq& \lim_n \int_0^\infty\frac{f_n(x+t)-f_n(x)}{t}dt\\ &\geq&\lim_n \int_{n+1-x}^\infty\frac{1}{t\:n}dt=\infty. \end{eqnarray*} On the other hand, we have $f(x)=\lim_n f_n(x)=0$ uniformly, such that $$\sup_{\gamma\in(0,2)}\int_0^\infty\frac{f(x+t)-f(x)}{t^\gamma}dt=0.$$

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  • $\begingroup$ Care to elaborate on the downvote? $\endgroup$ – mbe Mar 19 '17 at 15:18
  • $\begingroup$ I didn't downvote, but I think there is a problem with $$\lim_n \int_{n+1-x}^\infty\frac{1}{t\:n}dt=\infty.$$ $\endgroup$ – user428573 Mar 23 '17 at 11:31
  • $\begingroup$ It's the limit of a sequence of integrals, all of which don't converge since they are some tail of the integral of 1/t. $\endgroup$ – mbe Mar 23 '17 at 11:41
  • $\begingroup$ I see. Also, I'm not quite sure how you got there from the previous line or why $f_n \to 0$ uniformly. $\endgroup$ – user428573 Mar 23 '17 at 12:20
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Added later: There are many counterexamples. Take any smooth bounded function $g$ on $\mathbb R$ whose derivative is positive everywhere (for example, $g(t)=\arctan t$). Set $f_n(t) = g(t)/n.$ Then $f_n \to 0$ uniformly on $\mathbb R,$ and the hoped-for equality fails at every $x.$

Proof: Fix $x\in \mathbb R.$ Then there exists $\delta > 0$ such that

$$g(x+t)-g(x) > g'(x)t/2\,\,\text { for } 0<t<\delta.$$

Thus

$$\int_0^\infty \frac{f_n(x+t)-f_n(x)}{t^\gamma}dt \ge \frac{1}{n}\int_0^\delta \frac{g(x+t)-g(x)}{t^\gamma}dt \ge \frac{1}{n}\frac{g'(x)}{2}\int_0^\delta \frac{t}{t^\gamma}dt.$$

Now the last expression equals

$$\frac{1}{n}\frac{g'(x)}{2}\frac{\delta^{2-\gamma}}{2-\gamma}.$$

As $\gamma \to 2^-,$ this $\to \infty.$ Thus the supremum is $\infty$ for each $n.$ So as before, we have $\infty$ on the left and $0$ on the right, this time for every $x.$


Previous answer: It would appear that the result fails: Define $f_n(t) = \sqrt t/n.$ Then $f_n \to 0$ uniformly on compact subsets of $[0,\infty).$ Let $x=0.$ Then the integrals on the left are

$$ \frac{1}{n}\int_0^\infty \frac{\sqrt t}{t^\gamma}\, dt.$$

Now we take the $\sup$ over $\gamma \in (0,2).$ But notice $\gamma = 3/2$ makes each of these integrals $\infty.$ Thus the supremum is $\infty$ for each $n,$ and of course $\lim_{n\to \infty} \infty = \infty.$

On the right, each integral is just $0,$ so this appears to be a counterexample.

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  • $\begingroup$ What if we consider $$\lim_n \sup_{\gamma \in (0,2)}\int_0^\infty \frac{f_n(x+t)-f_n(x)}{e^{|t|}t^\gamma}dt?$$ Would the result hold then? $\endgroup$ – user428573 Mar 31 '17 at 22:13
  • $\begingroup$ @Riku No, because of the problem at $0.$ You would still end up with $\infty$ on the left, and $0$ on the right. $\endgroup$ – zhw. Mar 31 '17 at 23:04
  • $\begingroup$ That's right. Then what condition should suffice to make the result hold? $\endgroup$ – user428573 Mar 31 '17 at 23:28
  • $\begingroup$ Actually I should amend my answer above, since you are requiring bounded continuous functions. Define $g(t) = \sqrt t, 0\le t \le 1,$ $g(t) = 1, t> 1.$ Then set $f_n(t) = g(t)/n.$ Same outcome. $\endgroup$ – zhw. Mar 31 '17 at 23:48
  • $\begingroup$ Right. But suppose $f_n$ is $C^1$ bounded and with bounded derivative. Then we can say only that the limit function $f$ is bounded and continuous. But is then the problem solved and the statement true? $\endgroup$ – user428573 Apr 1 '17 at 8:42

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