Basic Forms
{Theorem 1.1}
Let (X,d) be a complete metric space and let f:X→R⋃{+∞} be a lsc function bounded from below. Suppose that ϵ>0 and z∈X satisfy
f(z)<infXf+ϵ
Then there exists
y∈X such that\
(i)
d(z,y)≤1,\
(ii)
f(y)+ϵd(z,y)≤f(z), and\
(iii)
f(x)+ϵd(x,y)≥f(y), for all
x∈X. \
\textbf{Proof}. Define a sequence
(zi) by induction starting with
z0:=z. Suppose that we have defined
zi. Set
Si:={x∈X|f(x)+ϵd(x,zi)≤f(zi)}
and consider two possible cases:(a)
infSif=f(zi). Then we define
zi+1=zi.(b)
infSif<f(zi). We choose
zi+1∈Si such that
f(zi+1)<infSif+12[f(zi)−infSif]=12[f(zi)+infSif]<f(zi)\eqno(1.1)
We show that
(zi) is a Cauchy sequence. In fact, if (a) ever happens then
zi is stationary for
i large. Otherwise,
ϵd(zi,zi+1)≤f(zi)−f(zi+1)\eqno(1.2)
Adding (1.2) up from
i to
j−1>i we have
ϵd(zi,zj)≤f(zi)−f(zj)\eqno(1.3)
Observe that the sequence
(f(zi)) is decreasing and bounded from below by
infXf, and therefore convergent. We conclude from (1.3) that
(zi) is Cauchy. Let
y:=limi→∞zi. We show that
y satisfies the conclusions of the theorem. Setting
i=0 in (1.3) we have
ϵd(z,zj)+f(zj)≤f(z)\eqno(1.4)
Taking limits as
j→∞ yields (ii). Since
f(z)−f(y)≤f(z)−infXf<ϵ, (i) follows from (ii). It remains to show that
y satisfies (iii). Fixing
i in (1.3) and taking limits as
j→∞ yields
y∈Si. That is to say
y∈⋂i=1∞Si
On the other hand, if
x∈⋂∞i=1Si then, for all
i=1,2,⋯,
ϵd(x,zi+1)≤f(zi+1)−f(x)≤f(zi+1)−infSif\eqno(1.5)
It follows from (1.1) that
f(zi+1)−infSif≤f(zi)−f(zi+1), and therefore
limi[f(zi+1)−infSif]=0. Taking limits in (1.5) as
i→∞ we have
ϵd(x,y)=0. It follows that
⋂i=1∞Si=y\eqno(1.6)
Notice that the sequence of sets
(Si) is nested, i.e., for any
i,
Si+1⊂Si. In fact, for any
x∈Si+1,
f(x)+ϵd(x,zi+1)≤f(zi+1) and
zi+1∈Si yields
f(x)+ϵd(x,zi)≤f(x)+ϵd(x,zi+1)+ϵd(zi,zi+1)≤f(zi+1)+ϵd(zi,zi+1)≤f(zi)(1.7)
which implies that
x∈Si. Now, for any
x≠y, it follows from (1.6) that when
i sufficiently large
x∉Si. Thus,
f(x)+ϵd(x,zi)≥f(zi). Taking limits as
i→∞ we arrive at (iii).
Other Forms
{Theorem 1.2}
Let (X,d) be a complete metric space and let f:X→R⋃{+∞} be a lsc function bounded from below. Suppose that ϵ>0 and z∈X satisfy
f(z)<infXf+ϵ
Then, for any
λ>0 there exists
y∈X such that\
(i)
d(z,y)≤λ,\
(ii)
f(y)+(ϵ/λ)d(z,y)≤f(z), and\
(iii)
f(x)+(ϵ/λ)d(x,y)≥f(y), for all
x∈X. \
{Theorem 1.3}
Let
(X,d) be a complete metric space and let
f:X→R⋃{+∞} be a lsc function bounded from below. Suppose that
ϵ>0 and
z∈X satisfy
f(z)<infXf+ϵ
Then there exists
y∈X such that\
(i)
d(z,y)≤ϵ√,\
(ii)
f(y)+ϵ√d(z,y)≤f(z), and\
(iii)
f(x)+ϵ√d(x,y)≥f(y), for all
x∈X. \
{Theorem 1.3}
Let
(X,d) be a complete metric space and let
f:X→R⋃{+∞} be a lsc function bounded from below. Then, for any
ϵ>0, there exists
y∈X such that
f(x)+ϵ√d(x,y)>f(y)
