Verify that if you have an image I with all zeroes except for a single location with value $1,$ that $($f $$I$)($a$,$b$)=$ f$.$ Try it for a $3$x$3$ filter f and a $5$x$5$ image I with a $1$ in the middle and zeros everywhere else. Note that the result $($f $$I$)($a$,$b$)$ might be a bit offset from f$($a$,$b$)$ depending on your indexing.