Integral inequality I want to bound in $\mathbb{R}^n$ this integral
$$
\int_{\lvert x-y\rvert<M} K(x,y) \left[\frac{1}{\lvert x-y\rvert^a}-\frac{1}{M^a}\right]dy
$$
with $K∈ L^1( \mathbb R^n \times \mathbb R^n)$.
My attempt
If I call $C=\sup_x \int K(x,y)dy$, 
$$
\int_{\lvert x-y\rvert<M} K(x,y) \left [ \frac{1}{\lvert x-y\rvert^a}-\frac{1}{M^a} \right ]dy \le C\int_{\lvert x-y\rvert<M} \left [ \frac{1}{\lvert x-y\rvert^a}-\frac{1}{M^a} \right ]dy
$$
Now: $\frac{1}{\lvert x-y\rvert^a}-\frac{1}{M^a}\le \frac{1}{M^a}$ in $D =\lbrace \lvert x-y \rvert <M \rbrace$ so
$$
C\int_D \left [ \frac{1}{\lvert x-y\rvert^a}-\frac{1}{M^a} \right ]dy \le C \left [ \int_D \frac{1}{\lvert x-y\rvert^a} -\int_D \frac{1}{M^a} \right ]
$$
First term is
$\int_D \frac{1}{\lvert x-y\rvert^a}$ and by polar coordinates is something like $\int_D \rho^{n-1-a}d \rho\le\frac{1}{M^{n-a}}$
while the second one is bounded by a costant $C_2$ so that (I think) last inequality is
$$
\int_{\lvert x-y\rvert<M} K(x,y) \left [ \frac{1}{\lvert x-y\rvert^a}-\frac{1}{M^a} \right ]dy \le C \frac{1}{M^{n-a}}
$$
 A: You don't seem to use the $x$ dependence anywhere so I'll go ahead and assume its just an indexing parameter, and that $K(x,·) ∈ L^1(\Bbb R^n)$ for each $x$ since you used this. Lets assume $a≠n$, otherwise there will be a log term instead of a power. Your steps are in the right direction but have mistakes. Defining $c(x) := ‖K(x,·)‖_{L^1}$, you get $$\left| ∫_{|x-y|<M} K(x,y) \left(\frac1{|x-y|^a} - \frac1{M^a}\right) \ \text dy \right|\leq c(x) \left| ∫_{|x-y|<M} \left(\frac1{|x-y|^a} - \frac1{M^a}\right) \ \text dy \right| $$
You omitted the absolute value bars. You can explicitly compute the integrals inside,
$$∫_{|x-y|<M} \frac1{|x-y|^a} \ \text dy = n V_n ∫_0^M \rho^{n-1-a} \ \text d\rho  = M^{n-a} = \frac{n V_n}{(n-a)M^{a-n}}$$
Note the exponent of $M$ here differs from yours. $V_n$ if I recall correctly is the volume of the unit ball on $\Bbb R^n$. Similarly
$$∫_{|x-y|<M} \frac1{M^a} \ \text dy = \frac{V_n}{M^{a-n}}$$
Which means that
$$ c(x) \left| ∫_{|x-y|<M} \left(\frac1{|x-y|^a} - \frac1{M^a}\right) \ \text dy \right| = \frac{c(x) V_n}{M^{a-n}}\left| \frac{n}{n-a} - 1\right| $$ 
