Image processing - Is mean filter shift-invariant? Update:
So, according to this article.
For a 2D mean filter,
$$
H[f(x,y)]=\frac{1}{NM}\sum_{k=0}^{N-1}\sum_{p=0}^{M-1}f(x-k,y-p)=g(x,y)\tag{1}
$$
$$
g(x+x_0,y+y_0)=\frac{1}{NM}\sum_{k=0}^{N-1}\sum_{p=0}^{M-1}f(x+x_0-k,y+y_0-p)\tag{2}
$$
$$
H[f(x+x_0,y+y_0)]=\frac{1}{NM}\sum_{k=0}^{N-1}\sum_{p=0}^{M-1}f(x+x_0-k,y+y_0-p)\tag{3}
$$
because (2) equals to (3), a 2D mean filter is shift-invariant.
Is the proof right?
BTW, if f(x,y) is a 2D image, such as

1 2 1
2 3 2
1 2 1


and g(x,y) is (using zero padding)

 8/9 11/9  8/9
11/9 15/9 11/9
 8/9 11/9  8/9


What does these
$$
H[f(x+x_0,y+y_0)]\tag{4}
$$
and
$$
g(x+x_0,y+y_0)\tag{5}
$$
mean?
(I want to prove it by using a image.)
Origin:
I read this article
and this Wikipedia page.
However, I still don't know what it means for (or how to apply to) a mean filter.
Can anyone give me an image example?
(for example, https://dsp.stackexchange.com/a/14435 is easier to understand)
The 1D mean filter defined by the following I/O relation is time/shift invariant;

$$y[n] = T\{x[n]\} = \frac{1}{d_1 + d_2 + 1} \sum_{k=n-d_1}^{n+d_2}{x[k]}$$

For $d_1$ and $d_2$ being positive integers, the filter computes the mean (or the average) of its input $x[n]$ in the range of its summation index. To show that it is shift invariant we shall simply show the truth of the following statement:  $y[n-d] = T\{x[n-d]\}$ for and $d$ and any input $x[n]$

$$ y[n-d] = \frac{1}{d_1 + d_2 + 1} \sum_{k=n-d-d_1}^{n-d+d_2}{x[k]} ~~~\overset{?}{=}~~~ T\{x[n-d]\} = \frac{1}{d_1 + d_2 + 1} \sum_{k=n-d_1}^{n+d_2}{x[k-d]}$$

Which is easily shown to be true by making the simple change of variable in the second sum as $k-d=k'$ and then optionally replacing $k'$ with $k$ again:
\begin{align}
T\{x[n-d]\} &amp;= \frac{1}{d_1 + d_2 + 1} \sum_{k=n-d_1}^{n+d_2}{x[k-d]} ~~~~ &amp;\scriptstyle{\text{replace k-d with k'}}\\
T\{x[n-d]\} &amp;= \frac{1}{d_1 + d_2 + 1} \sum_{k'=n-d-d_1}^{n-d+d_2}{x[k']} &amp;\scriptstyle{\text{replace k' with k again}}\\
T\{x[n-d]\} &amp;= \frac{1}{d_1 + d_2 + 1} \sum_{k=n-d-d_1}^{n-d+d_2}{x[k]} &amp;\scriptstyle{\text{recognise that this is =  y[n-d]}}\\
T\{x[n-d]\} &amp;= y[n-d] &amp;\scriptstyle{\text{Hence the given System is LTI}}\\
\end{align}

This is extended to 2D in a straight-forward manner