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?

 A: 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.
A: 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. 
