How to simplify multiple addition and convolution operations into one convolution kernel I need to perform such a conversion to simplify my image processing problem (sharpening, in green are the knowns, in red the unknowns):

\begin{align}
y(n,m) &amp;= \color{green}{x(n,m)} * \left[ \color{red}{f_1(n,m) + f_2(n,m)} + \color{green}{f_3(n,m)}\right]\\
&amp;= \color{green}{x(n,m) + kx(n,m) - x(n,m) * kg(n,m)}
\end{align}

Notice that the convolution done here is two-dimensional. And I know that $$ f_3  = -kg ;$$
where $k$ is a constant. However, I could not find $f_1$ and $f_2$, I tried to use convolution Theorem to switch to FT then going back to time/spatial-domain but it did not  obtain the same result. 
Here is a better explanation for the requested problem: 

To perform such a process to an input image, instead of performing the three separated steps shown in the top figure, I need to find such a kernel; let's call f where 
$$
f = f_1 + f_2 + f_3
$$
and then convolve it directly with the input image to find the same output image described in the block diagram figure. 
Any help/idea?
Let's rearrange
\begin{align}
x(n,m) * \left[ f_1(n,m) + f_2(n,m) + f_3(n,m)\right]
&amp;= x(n,m) + kx(n,m) - x(n,m) * kg(n,m)\\
x(n,m) * \left[ f_1(n,m) + f_2(n,m) \right]
&amp;= x(n,m) + kx(n,m)\\
&amp;=(1+k)x(n,m)
\end{align}
It directly follows from the linearity of convolution that $f_1+f_2=(1+k)\delta_{n,m}^{(N \times M)}$, where $\delta^{(N\times M)}$ is the Kronecker delta of the same size as your input $x$, i.e. an all-zero matrix with a 1 in the $0,0$ element; the neutral element for 2D-convolution.

So, it's possible to calculate what the sum of $f_1 + f_2$ is, but not the individual $f_1$ and $f_2$; $f_1= \delta^{(N\times M)}, f_2 = k\delta^{(N\times M)}$ is as correct as
$f_1=\begin{pmatrix}1234 &amp; 1 &amp; \cdots &amp;1\\0 &amp; 0 &amp; \cdots &amp;0\\0 &amp; 0 &amp; \cdots &amp;0\\\end{pmatrix},f_2=\begin{pmatrix}-1233+k &amp; -1 &amp; \cdots &amp;-1\\0 &amp; 0 &amp; \cdots &amp;0\\0 &amp; 0 &amp; \cdots &amp;0\\\end{pmatrix}$.