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