Proof
Let
for
Consider the
problems


(Penalized problem)

where
and
is such that
By the classic version of the Weierstrass theorem there exists a solution
of the problem (
Penalized problem
) for every
.
In
particular,
Note that
and
.
Hence, we rewrite the last inequality
as


(FJ proof 1)

By construction,
is a bounded sequence. Therefore, is has one or more limit points
.
The
is smooth, hence,
is bounded.
Therefore,
because otherwise
and
cannot be bounded by the
.
Thus, all the limit points
are feasible:
Therefore, by the construction of
and
,


(FJ proof 2)

By passing to the limit the inequality (
FJ proof 1
)
and combining with (
FJ proof 2
)we
conclude
for every limit point
.
Thus
is convergent and
According to the proposition (
Minimum
of a smooth
function
)
By convergence
,
for large enough
the
is inside
,
hence
We restrict our attention to such
.
We
calculate


(FJ proof 3)

and introduce the
notation
Note that the sequence of
is
bounded:
Hence, it has a limit point
By dividing (
FJ proof 3
) with
we obtain
We pass the last relationship to the limit
and arrive
to
(compare with the definition (
Normal cone
)).
To see that the statement (4) holds, note that by construction of
,
,
if
then
for large enough
.
If
then the
th
component of
:
has to vanish as
.
Hence, if
then
vanishes quicker than any of the
for
.
The consideration for
is identical.