Solutions to Variational Analysis by Rockafellar and Wets

Solutions to Variational Analysis by Rockafellar and
Wets

Robert Bassett
May 14, 2015

A quick test

1 1D

1.12 A function f : Rn Ñ R is continuous if and only if it is both lower semicontinuous and
upper semicontinuous.

By the definition of the lower limit

„

lim inf f pxq “ lim inf f pxq .
xÑx δÑ0 xPBx,δ

Using the definition of the limit (see, for example Bartle and Sherbert’s Introduction
to Real Analysis) this means

ˇˇ

@ Dδ0 ą 0 such that δ ă δ0 ñ ˇ inf f pxq ´ lim inf f pxqˇˇ ă .
ˇ

ˇxPBpx,δq xÑx ˇ

Note that infxPBpx,δq f pxq is increasing in δ so that this last expression is equivalent to

lim inf f pxq ´ inf f pxq ă .
xÑx xPBpx,δq

Similarly, the definition of upper limit means

@ Dδ1 ą 0 such that δ ă δ1 ñ sup f pxq ´ f pxq ă .

xPBpx,δq

Now we will complete the problem. Assume that f is both lower and upper semicontin-
uous. Note that each of the implications that follow are reversible, so that the reverse
direction also follows.

1

To show that f is continuous at x, we need to show that @ ą 0Dδ2 such that x P
Bpx, δq ñ |f pxq ´ f pxq| ă . Fix ą 0. Take δ “ mintδ0, δ1u, where delta0 and
δ1 come from the definitions of lower and upper semicontinuity. We then have, for
x P Bpx, δq,

|f pxq ´ f pxq|

“ maxtf pxq ´ f pxq, f pxq ´ f pxqu

ď maxt sup f pxq ´ f pxq, sup f pxq ´ f pxqu

xPBpx,δq xPBpx,δq

maxt sup f pxq ´ f pxq, f px ´ inf f pxqu
xPBpx,δq
xPBpx,δq

ă maxt , u “ .

So we are done. Note that splitting the absolute value into the maximum of its positive
and negative parts is the crux of this problem, and allows us to incorporate the def-
initions of lower and upper semicontinuity. So“controlling” both of these expressions
gives us continuity.

1.13 For an arbitrary function f : Rn Ñ R and a pair of elements x P Rn and α P R, one
has

a px, αq P clpepif q if and only if α ě lim infxÑx f pxq.

Assume px, αq P clpepif q. Then there is a sequence pxν, ανq Ñ pxαq with αν ě
f pxνq for all ν P N. Taking lim infνÑ8 of both sides of this inequality gives

α ě lim inf f pxνq
nÑ8

this last expression is greater than or equal to lim infxÑx f pxq by Lemma 1.17

For the reverse direction, if α ě lim infxÑx f pxq, let xν by a sequence converging
to x such that limnÑ8 f pxνq Ñ lim infxÑx. Such a sequence exists by Lemma
1.17. Then consider the sequence pxν, f pxνq ` α ´ lim infxÑx f pxqq. Since α ě
lim infxÑx f pxq, this is in epif . This clearly converges to px, αq, so that px, αq P
clpepif q.

b px, αq P intpepif q if and only if α ą lim supxÑx.

For ease of computation in this problem, we’ll equip Rn`1 with the one-norm
topology. In other words,

n`1

ÿ
dpx, yq “ |xi ´ yi| .

i“1

Because Rn`1 is finite dimensional, all norms on it are equivalent and hence we
have this liberty. See, for example, Applied Analysis by Hunter Nachtergale,
section 5.4.

2

We will prove the negation of this statement. Assume that α ď lim supxÑx f pxq.
Take xν Ñ x such that

lim f pxνq “ lim sup f pxq.
νÑ8
xÑx

Such a sequence exists by Lemma 1.7 applied to lim sup. We need to show that

px, αq R intpepif q, i.e. for every ą 0, Bppx, αq, q Ę epipf q. Fix ą 0. Choose N

so that ν ą N implies

dppxν, f pxνq ´ plim sup f pxq ´ αqq, px, αqq ă {2.

xÑx

This can be done because by our choice of xν, f pxνq Ñ lim supxÑx f pxq.

Consider the sequence

pxν, f pxνq ´ plim sup f pxq ´ αq ´ {2q.

xÑx

This is not in epif because α ď lim supxÑx f pxq. Then, because we work with the
1-norm, for ν ą N ,

dppxν, f pxνq ´ plim sup f pxq ´ αq ´ {2q, px, αqq

xÑx

ˇˇ

ď ||xν ´ x||1 ` ˇˇf pxνq ´ plim sup f pxq ´ αq ´ α ´ {2ˇˇ
ˇ ˇ
x´x

ˇˇ

ď ||xν ´ x||1 ` ˇˇf pxν ´ lim sup f pxqˇˇ ` {2
ˇ ˇ
xÑx

“ dppxnu, f pxνq ´ plim sup f pxq ´ αqq, px, αqq ` {2

xÑx

ă.

So for ν ą N , pxν, f pxνq ´ plim supxÑx f pxq ´ αq ´ {2q is in the ball of px, αq,
but not in the epigraph. We conclude that px, αq R intpepif q.

For the reverse direction, assume px, αq R intpepif q. Then for every ą 0 there
exists px, αq such that px, αq P Bppx, αq, q and not in epif . Build a sequence by
taking “ 1{ν and choosing a corresponding pxν, ανq. Then since each of these
points are not in the epigraph of f ,

αν ă f pxνq

Taking limits of this expression

alpha ď lim f pxνq ď lim sup f pxq
νÑ8
xÑx

where the last inequality follows from the characterization of lim sup as the max-
imum limit point (see Lemma 1.7 for lim sup).

3

c px, αq R clpepif q if and only if px, αq P intphypof q.

Note that px, αq P epi ´ f if and only if px, ´αq P hypof . This follows from
opening up the definitions and the fact that

α ě lim inf ´f pxq ðñ ´α ď lim sup f pxq.
xÑx
xÑx

Thus we have

px, αq R clpepif q lðo mño n α ă lim inf f pxq
xÑx

part (a)

ðñ ´α ą lim sup ´f pxq

xÑx

lðomñonpx, ´αq P intpepi ´ f q

part (b)

lðomñon px, αq P hypof

comment above

d px, αq R intpepif q if and only if px, αq P clphypof q.
Similiar to (c), we have

px, αq R intpepif q lðomñon α ď lim sup f pxq
xÑx
by (b)

ðñ ´α ě lim inf ´f pxq
xÑx

lðomñonpx, αq P clpepi ´ f q

by (a)

ðñ px, αq P clphypof q.

1.14

4