Local nature of a surface charge density Boundary $S$ of a cavity in a very large (perfect) conductor is a connected compact (smooth) surface. A positive point charge $+q$ is placed inside this cavity. From Gauss' law we know that the total induced charge on the surface $S$  is $-q$. However, is the surface charge density on $S$ necessary locally negative (at every point of $S$)? How to prove (or disprove) this assertion?
I'm looking for a precise, robust argument, not hand-waving type of answers.
 A: Assume the contrary,suppose a point exists such that the local charge density is positive,say point A.
Now from Gauss' law the total charge on the inner surface is negative.So there must exist a point B at which the local charge density is negative(otherwise the net charge will be positive).

Now consider the field line from point A.It will originate from the conductor and it can have two possibilities:
1)It meets the given charge:not possible as the given charge is positive.
2)It meets at some point on the conductor.
Since the first case is impossible,we have to accept the second one.The blue curve is the field line.
Now consider $\oint \vec E \cdot \vec dx$ over the coloured loop(blue and orange ) as shown in the figure.We can divide this integral in two parts:
1)A to B inside the conductor(Blue curve):since this is a field line(according to the second possibility) the integral is some non-zero real number.
2)B to A inside the conductor(Orange curve):since the field is zero,the integral is zero.
So finally,we have

$\oint \vec E \cdot \vec dx=$non-zero real number

which at the face of it contradicts the fact the the E field is conservative.Contradiction.
Done!
