Suppose $f\in C^\infty(\mathbb{R})$ is compactly supported on $[-N,N]$, and $K\in L^1(\mathbb{R})$ with $\int_\mathbb{R}K(x)dx=1$. Define $$K_t(x)=\dfrac{1}{t}K\left(\dfrac{x}{t}\right)$$
It follows easily that $\int_\mathbb{R}K_t(x)=\int_\mathbb{R}K(x)dx=1$.
I'm in the process of proving some convergence bound, and I wonder whether we can show that $$\lim_{t\rightarrow 0}\left(\sup_{|x|\leq N+1}\left|\int_{-\infty}^\infty(f(x-y)-f(x))K_t(y)dy\right|\right)=0 $$
Let $\epsilon >0$. Then there is $M>0$ such that $\int_{|x|\geq M} |K| < \epsilon$ as $K$ is in $L^1$. That is
$$ \int_{|x| \geq tM} |K_t| < \epsilon\ .$$
Write
$$\int_\mathbb{R} \big(f(x-y) - f(x)\big) K_t(y) dy = \int_{|y|< tM} + \int_{|y|\geq tM}\big(f(x-y) - f(x)\big) K_t(y) dy$$
As $f$ is uniformly continuous, there is $\delta >0$ such that
$$|f(x-y) - f(x)|< \epsilon,\ \ \forall x\in \mathbb R,\ \ \forall |y| < \delta\ .$$
In particular, if $t < \delta /M$, then
$$\bigg|\int_{|y| < tM}\big(f(x-y) - f(x)\big) K_t(y) dy\bigg| < \epsilon \ ||K||_1$$
while the second term is bounded by
$$2||f||_0 \int_{|y|\geq tM} |K_t| < 2||f||_0\epsilon\ ,$$
Thus
$$\bigg| \int_\mathbb{R} \big(f(x-y) - f(x)\big) K_t(y) dy \bigg| < (2||f||_0 + ||K||_1) \epsilon$$
for all $x$ and for all $t \leq \delta /M$. Hence
$$\lim_{t\to 0} \bigg(\sup_x \bigg|\int_\mathbb{R} \big(f(x-y) - f(x)\big) K_t(y) dy\bigg|\bigg) = 0\ .$$
Note that we need only that $f\in C^0_c(\mathbb{R})$ and $\int K =1$ is not used.