1 Functional analysis tools 1. Banach space: weak topologies Let X , Y be Banach spaces (normed vector spaces in which every Cauchy se- quence has a limit). Let A be linear X ! Y . Then A is continuous i↵ there exists c> 0 such that (1.1) kAxk Y ckxk X , for all x 2 X . The space L(X, Y ) of linear mappings X ! Y , is a Banach space when endowed with the norm (1.2) kAk L(X,Y ) := sup{kAxk Y ; kxk X 1}. 1.0.1. Unique extension of linear mappings. Let X , Y be Banach spaces, and E be a dense subspace of X . Let A : E ! Y be linear and such that for some c> 0: (1.3) kAek Y ckek X , for all e 2 E. Then A has a unique extension to L(X, Y ), i.e. there exists a unique A 2 L(X, Y ), such that Ae = Ae, for all e 2 E, and for x 2 X (the limit below exists): (1.4) Ax := lim k {Ae k , e k 2 E, e k ! x}. 0 Chapter from the lecture notes on ’Optimal control of partial di↵erential equations’ by J.F. Bonnans. Version of April 11, 2020. Updates and additional material on the page http://www.cmap.polytechnique.fr/⇠bonnans/notes/cedp/cedp.html 9
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1Functional analysis tools
1. Banach space: weak topologies
Let X, Y be Banach spaces (normed vector spaces in which every Cauchy se-
quence has a limit).
Let A be linear X ! Y . Then A is continuous i↵ there exists c > 0 such that
(1.1) kAxkY ckxkX , for all x 2 X.
The space L(X, Y ) of linear mappings X ! Y , is a Banach space when endowed
with the norm
(1.2) kAkL(X,Y ) := sup{kAxkY ; kxkX 1}.1.0.1. Unique extension of linear mappings. Let X, Y be Banach spaces, and E
be a dense subspace of X.
Let A : E ! Y be linear and such that for some c > 0:
(1.3) kAekY ckekX , for all e 2 E.
Then A has a unique extension to L(X, Y ), i.e. there exists a uniqueA 2 L(X, Y ),
such that Ae = Ae, for all e 2 E, and for x 2 X (the limit below exists):
(1.4) Ax := lim
k{Aek, ek 2 E, ek ! x}.
0Chapter from the lecture notes on ’Optimal control of partial di↵erential equations’ by J.F.Bonnans. Version of April 11, 2020. Updates and additional material on the page
1.0.2. Multilinear mappings. Let X := X1 ⇥ · · · ⇥ Xn, a : X ! Y multilinear,
the Xi and Y being Banach spaces. Then a is continuous i↵ there exists c > 0
such that
(1.5) ka(x1, . . . , xn)kY ckx1kX1 · · · kxnkXn
for all x = (x1, . . . , xn) 2 X.
If E is a dense subset of X, a : E ! Y is multilinear, and the above inequality holds
for all x 2 E, then a has a unique continuous multilinear extension to X.
Remark 1.1. If the above multilinear mapping a is continuous, then it is of
class C1, with derivative at x 2 X in direction h 2 X given by
(1.6) Da(x)h =
nX
i=1
a(x1, . . . , xi�1, hi, xi+1, . . . , xn).
Example 1.2. Let p, q, r belong to (1,1), such that 1/r = 1/p + 1/q, and let
⌦ be an open subset of Rn. By Holder’s inequality, if f 2 Lp
(⌦) and g 2 Lq(⌦),
then fg 2 Lr(⌦) and
(1.7) kfgkLr(⌦) kfkLp(⌦)kgkLp(⌦).
It follows that the mapping Lp(⌦)⇥ Lq
(⌦) ! Lr(⌦), (f, g) 7! fg is of class C1
.
1.0.3. Topological dual. Let X be a Banach space. A linear form over X is a
linear mapping X ! R.We call topological dual (or, in short, dual) and denote by X⇤
, the set of
continuous linear forms over X.
The action (duality product) of x⇤ 2 X⇤over x 2 X is denoted by hx⇤, xiX . The
dual X⇤is a Banach space, endowed with the dual norm
(1.8) kx⇤kX⇤:= sup{|hx⇤, xiX |; kxkX 1}.
1.0.4. Bidual; reflexive spaces. The bidual (dual of the dual) of X is denoted by
X⇤⇤.
With x 2 X associate the linear form over X⇤:
(1.9) `x(x⇤) := hx⇤, xiX .
The mapping x ! X⇤⇤, x 7! `x is isometric: k`xkX⇤⇤
= kxkX .So, we can identify X with the image of `, which is a closed subspace of X⇤⇤
.
We say that X is reflexive if ` is onto: in that case we can identify X and X⇤⇤,
i.e., X is the dual of X⇤.
1.0.5. Transpose operator. Given A 2 L(X, Y ), fix y⇤ 2 Y ⇤. The mapping
(1.10) `y⇤ : X ! R, x 7! hy⇤, AxiYis continuous since |`y⇤(x)| ky⇤kkAkkxk, so that
(1.11) k`y⇤kX⇤ ky⇤kkAk.
1. BANACH SPACE: WEAK TOPOLOGIES 11
Also y⇤ 7! `y⇤ is linear and (by the above inequality) continuous: so, we may denote
it as A>y⇤, with A> 2 L(Y ⇤, X⇤), and:
(1.12) hA>y⇤, xiX = hy⇤, AxiY , for all x 2 X, y⇤ 2 Y ⇤.
1.0.6. Examples of Banach spaces.
• Hilbert space X: scalar product denoted by (·, ·)X .Associated norm kxkX := (x, x)1/2X . Such spaces are reflexive.
Riesz theorem: if x⇤ 2 X⇤, there exists y 2 X such that
hx⇤, xiX = (y, x)X , for all x 2 X.
• ⌦ open subset of Rn, Lp
(⌦) reflexive if p 2 (1,1) and
(1.13) Lp(⌦)
⇤= Lq
(⌦), 1/p+ 1/q = 1, p 2 [1,1),
• ⌦ as above: L1(⌦) and L1
(⌦) not reflexive.
1.0.7. Punctual convergence by a density argument.
Lemma 1.3. Let X, Y be Banach spaces and Ak be a bounded sequence inL(X, Y ). Let Akx ! 0 for all x in a dense subset E of X. Then Akx ! 0,for all x 2 X.
Proof. Let x 2 X. Fix " > 0, and e 2 E such that
(1.14) sup
`kA`kkx� ek ".
Then
(1.15) kAkxk kAk(x� e)k+ kAkek sup
`kA`kkx� ek+ kAkek.
It follows that
(1.16) lim sup
kkAkxk ".
The result follows. ⇤1.0.8. Weak convergence. Given a sequence {xk} in a Banach space X, and
x 2 X, we say that xk weakly converges to x, and write xk * x if
(1.17) hx⇤, xkiX ! hx⇤, xiX , for all x⇤ 2 X⇤.
Lemma 1.4. Any weakly convergent sequence is bounded and, if X is reflexive,any bounded sequence has a weakly convergent subsequence.
Proof. See Brezis [10]. ⇤1.0.9. Testing over a dense subset.
Lemma 1.5. We have that xk * x in the Banach space X i↵ xk is bounded and,for some dense subset E of X⇤:
(1.18) hx⇤, xkiX ! hx⇤, xiX , for all x⇤ 2 E.
Proof. Apply lemma 1.3. ⇤
12 1. FUNCTIONAL ANALYSIS TOOLS
1.0.10. About unbounded sequences. A counterexample is as follows. In X = `2
(space of summable square sequence) identified with its dual, let xk := kek (ek is
the kth element of natural basis) and E be the subspace of elements with finitely
many nonzero coordinates.
The hypothesis (1.18) of the lemma holds with x = 0, but xk does not weakly
converge since it is unbounded. A direct argument for the impossibility of weak
convergence is: let z :=
Pk ek/k
2/3, then z 2 X and (z, xk)X = k1/3
is unbounded.
1.0.11. Transportation of weak convergence by linear operators. If A 2 L(X, Y )
Sequence of constant norm, weakly (but not strongly) converging to 0.
Remark 1.7. In the case of a Lpspace, see the related Brezis-Lieb theorem 1.17.
1.0.13. Weak⇤ convergence. Let X be a Banach space. We say that the sequence
x⇤k in X⇤
weakly⇤ converges to x⇤ 2 X⇤if
(1.23) hx⇤k, xi ! hx⇤, xi, for all x 2 X.
Note that, if X is reflexive, then the weak and weak⇤ convergence coincide. In
general, we can only say that the former implies the later. Then, see Brezis [10],ch. 3, Coro. 3.30:
Lemma 1.8. Let X be a separable Banach space (i.e., there exists a dense se-quence). Then any bounded sequence in X⇤ has a weakly⇤ converging subsequence.
1. BANACH SPACE: WEAK TOPOLOGIES 13
Proof. Let x⇤k be a bounded sequence in X⇤
and x` be a dense sequence in X.
It su�ces to check that there exists x⇤ 2 X⇤such that up to the extraction of a
subsequence of x⇤k:
(1.24) hx⇤k, x`i ! hx⇤, x`i, for all ` 2 N .
We obtain this by the following diagonal extraction argument. Indeed, since
hx⇤k, x0i is bounded there exists a subsequence k0
i , for i 2 N, and ↵0 2 R such that
(1.25) |hx⇤k0i
, x0i � ↵0| 2
�i.
By induction we can construct, for any j 2 N⇤, a subsequence i 7! kji of kj�1
i such
that
(1.26) |hx⇤kji
, xii � ↵j| 2
�i.
Then the sequence j 7! x⇤j := x⇤
kjj
satisfies
(1.27) hx⇤j , x`i ! ↵`, for all ` 2 N.
For x 2 X, define
(1.28) hx⇤, xi := lim
jhx⇤, x`
j
i,
where `j is a subsequence such that x`j
! x. One can easily check that the limit
does not depend on the subsequence, and that this defines a continuous linear form
that is a weak⇤ limit of a subsequence of x⇤k. ⇤
Example 1.9. Let ⌦ be an open subset of Rn. Then X = L1
(⌦) is separable.
So, any bounded sequence in X⇤= L1
(⌦) has a weakly⇤ converging subsequence.
1.0.14. Closed convex sets are weakly sequentially closed. Indeed, let X be a
Banach space and K ⇢ X be convex and closed. Let {xk} ⇢ K weakly converge to
x. We need to prove that x 2 K.
If this is not true, by the Hahn Banach theorem (see Brezis [10], ch.1), thereexists x⇤ 2 X⇤ strictly separating x from K:
(1.29) hx⇤, xi < inf{hx⇤, xi; x 2 K}.then we get a contradiction since
(1.30) hx⇤, xi < lim
khx⇤, xki = hx⇤, xi.
1.0.15. What about weakly⇤ closed convex sets ? While for dual spaces we can-
not use the above separation argument, in practice we often find sets which are
intersection of “weakly⇤ closed” halfspaces, i.e., for some E ⇢ X ⇥ R:
(1.31) K := {x⇤ 2 X⇤; hx⇤, xiX ↵, for all (x,↵) 2 E}.
Clearly such a set is weak⇤ sequentially closed.
14 1. FUNCTIONAL ANALYSIS TOOLS
Example 1.10. Let X = L1(Rn
), f 2 X and f ⇤ 2 X⇤= L1
(Rn). We say that
f � 0 if f(x) � 0 a.e., and that f ⇤ � 0 if hf ⇤, fiX � 0, for all f � 0. Thus, a weak⇤limit of a nonnegative sequence in X⇤
is nonnegative.
1.0.16. Case of probability measures. Let ⌦ ⇢ Rnbe compact, and X = C(⌦)
be the Banach space of continuous functions over ⌦, endowed with the norm
(1.32) kfk := max{|f(x)|; x 2 ⌦}.The dual space is X⇤
:= M(⌦), space of bounded Borel measures over ⌦.
The set P(⌦) of Borel probability measures over ⌦ is a sequentially weakly⇤ closed
subset of M(⌦).
Since X is separable it follows that a (necessarily) bounded sequence of prob-
ability measures over ⌦ has a subsequence, weakly⇤ converging to a probability
measure.
Remark 1.11. The set ⌦ being as before, we can identify L1(⌦) with the subset
of M(⌦) of measures having a density. So, a bounded sequence in L1(⌦) will have a
weakly⇤ converging subsequence in M(⌦), but not necessarily a weakly converging
subsequence in L1(⌦).
1.0.17. Jensen’s inequality. Let � : R ! R [ {+1} be convex and lowersemi continuous (l.s.c.):
(1.33) �(x) lim inf
x!x�(x), for all x 2 R.
Let X(!) be a scalar random variable over the probability space (⌦,F , µ), withfinite expectation. Then
(1.34) �(IEX) IE�(X).
In particular if �(x) = |x|p, 1 p < 1:
(1.35) |IEX|p IE|X|p = kXkpLp(⌦).
2. Integration
2.1. Classical results. Three basic theorems
(1) Monotone convergence
(2) Dominated convergence
(3) Fatou lemma
2.1.1. Vitali’s theorem.
Definition 1.12 (Uniform integrability). Let (⌦,F ,P) be a probability space.
We say that a set E of measurable functions is uniformly integrable if, for all
" > 0, there exists M" > 0 such that
(1.36) IE|f |1{|f |>M"
} ", for all f 2 E.
2. INTEGRATION 15
Theorem 1.13. Let (⌦,F ,P) be a probability space, and fk be a uniformly inte-grable sequence in L1
(⌦), with a.s. finite limit f . Then f 2 L1(⌦), and fk ! f in
L1(⌦).
Remark 1.14. The uniform integrability of the sequence fk means that for all
" > 0, there exists M" > 0 such that
(1.37) IE|fk|1{|fk
|>M"
} ", for all k 2 N.Observe that this hypothesis is weaker than the one for the dominated convergence
theorem.
Counter example: ⌦ = R, Lebesgue measure, fk(x) := f(x+k) where f nonzero,
continuous with compact support. So, in general the conclusion does not hold when
meas(⌦) = 1.
2.2. Weak and a.e. convergence.
Lemma 1.15. Let ⌦ be an open subset of Rn, and for 1 < q < 1, a boundedsequence gk in Lq
(⌦), converging a.e. to some g. Then g 2 Lq(⌦) and gk weakly
converges to g in Lq(⌦).
Proof. We have that g 2 Lq(⌦), since by Fatou’s lemma,
(1.38) kgkqLq(⌦) =
Z
⌦
|g(!)|qd! lim inf
k
Z
⌦
|gk(!)|qd!.
For any nonzero N 2 N, set ⌦N := ⌦ \ B(0, 1/N), and
(1.39) EN := {x 2 ⌦N ; |gk(x)� g(x)| 1, for all k � N}.Let h 2 Lp
Extending v, g by 0 over Rn \ ⌦, we see that v ��v = g over Rn, and so:
(1.111) kvijk2 kgk2 c'�kfk2 + kukH1(⌦)
�.
Let ⌦
0 ⇢ ⌦ be open, with compact closure in ⌦. Choosing ' with value 1 on ⌦
0, so
that u = v on ⌦
0, deduce that
(1.112) kuijkL2(⌦0) c',⌦0�kfk2 + kukH1(⌦)
�.
Remark 1.36. Typically, by variational methods, we can estimate kukH1(⌦"
),
under some boundary conditions. So, the above estimate makes sense.
Remark 1.37. Obtaining estimates near the boundary depends on the boundary
conditions and is much more involved, see e.g. Lions and Magenes [23].
6.2. Complement: strong solutions in Lp spaces.
26 1. FUNCTIONAL ANALYSIS TOOLS
6.2.1. The Poisson equation in Lp spaces. The Fourier approach is the key for
obtaining the following extension (see the hard proofs in Gilbarg and Trudinger
[17]):
Theorem 1.38. Let p 2 (1,1) and f 2 Lp(Rn
). Then the equation
(1.113) u��u = f in Rn
has a unique solution in W 2,p(Rn
), and for some cp > 0 not depending on f :
(1.114) kukW 2,p(Rn) cpkfkLp(Rn).
6.2.2. Fractional Sobolev space. Following Grisvard [18], for s = m + �, withm 2 N and � 2 (0, 1), and p 2 [1,1], set
(1.115) W s,p(Rn
) := {u 2 Wm,p(Rn
); Ns,p(u) < 1},where
(1.116) Ns,p(u) :=X
|↵|=m
Z
Rn
Z
Rn
D↵u(x)�D↵u(y)|p|x� y|n+�p
dxdy.
This space is endowed with the norm
(1.117) kukW s,p(Rn) :=
⇣kukpWm,p(Rn) +Ns,p(u)
⌘1/p
.
This is a Banach space (a Hilbert space if p = 2).
6.2.3. Lp spaces based on the Fourier transform. For s 2 R+, let Gs denote the
Bessel potential of order s, defined by its Fourier transform
(1.118) FGs(⇠) = (1 + |⇠|2)s/2.For p 2 (1,1), define
(1.119) Hs,p(Rn
) := {u 2 L1loc(Rn
); Gs ⇤ u 2 Lp(Rn
)}.The following holds, see Grisvard [18, Ch. 1]: Hs,p
(Rn) = W s,p
(Rn) if s is integer,
and also,
(1.120) W s0,p(Rn
) ⇢ Hs,p(Rn
) ⇢ W s00,p(Rn
), whenever s0 > s > s00.
7. Elliptic control over Rn
7.1. Linear-quadratic setting.
(1.121) y(x)��y(x) = f(x) + u(x), x 2 Rn.
with f 2 L2(Rn
) given and u 2 L2(Rn
). Cost function
(1.122) J(u, y) := 12
Z
R(y(x)� yd(x))
2dx+
12
Z
Ru(x)2dx.
Optimal control problem
(P ) Min
u,yJ(u, y) s.t. (1.121); u 2 KU ,
7. ELLIPTIC CONTROL OVER Rn 27
with KU nonempty closed convex subset of U := L2(Rn
). State space Y := H2(Rn
).
For each u 2 U , the state equation (1.121) has a unique solution denoted by y[u] inY , and u 7! y[u] is a�ne and continuous U ! Y . The reduced cost function is
(1.123) F (u) := J(u, y[u]).
Being continuous and strongly convex, it has a unique minimizer u over KU , and
each minimizing sequence uk in KU strongly converges to u.Reduction Lagrangian: sum of cost function and product of state equation
by a multiplier:
(1.124) L(u, y, p := J(u, y) +
Z
Rn
p(x) (f(x) + u(x) +�y(x)� y(x)) dx.
A priori the space for the state equation is L2(Rn
), so we should take the multiplier
in the dual space L2(Rn
), but we decide to search for it in the smaller space Y . Then
by Green’s theorem over Y ⇥ Y (valid over the dense subset D(Rn)
2and extended
by continuity):
(1.125)
Z
Rn
p(x)�y(x)dx =
Z
Rn
y(x)�p(x)dx,
so that for y = y[u] and z 2 Y :
(1.126) Lyz =
Z
Rn
(y(x)� yd(x) +�p(x)� p(x))z(x)dx.
For y = y[u], this gives the costate equation, with unique solution p[u] in Y :
(1.127) p(x)��p(x) = y(x)� yd(x), x 2 Rn.
The linearized state equation, with (v, z) 2 U ⇥ Y , is:
(1.128) z(x)��z(x) = v(x), x 2 Rn.
And we get as expected, writing z = z[v] and p = p[v]:
(1.129)
F 0(u)v =
Z
Rn
((y(x)� yd(x))z(x) + u(x)v(x))dx,
=
Z
Rn
((p(x)��p(x))z(x) + u(x)v(x))dx,
=
Z
Rn
((z(x)��z(x))p(x) + u(x)v(x))dx,
=
Z
Rn
(p(x) + u(x))v(x)dx.
Since U = L2(⌦), this means that the derivative of F is
(1.130) F 0(u) = p[u] + u.
28 1. FUNCTIONAL ANALYSIS TOOLS
Let u be solution of the optimal control problem. Set y = y[u], p = p[u]. if F is
strictly convex, (u, y, p) is the unique solution in U⇥Y ⇥Y of the optimality system
(1.131)
8>><
>>:
y ��y = f + u;
p��p = y � yd;Z
Rn
(p(x) + u(x))(v(x)� u(x))dx � 0, for all v 2 KU .
The above inequality is equivalent to u = PKU
(�p). Eliminating u, we obtain that
(y, p) is the unique solution in Y ⇥ Y of
(1.132)
⇢y ��y = f + PK
U
(�p);p��p = y � yd.
7.2. Case when KU = U . Then u = �p and (y, p) is unique solution in Y ⇥Yof:
(1.133)
⇢y ��y = f � p;p��p = y � yd.
This is a system of two coupled linear elliptic equations. We next see how
to study this system in a direct way, i.e., without reference to the optimal control
problem.
7.2.1. Direct Study based on the Fourier transform. Set µ := 1 + 4⇡2|⇠|2. Ap-
plying the Fourier transform to the above system we find that
(1.134) µy + p =
ˆf ; �y + µp = yd.
We find that the unique solution satisfies
(1.135) (1 + µ2)y = µ ˆf � yd; (1 + µ2
)p =
ˆf + µyd.
Therefore (µy, µp) 2 L2(Rn
), so that (as expected) y and p belong to Y .
7.2.2. Direct study based on the Lax-Milgram theorem. Here we anticipate on the
Lax-Milgram theory, presented in the next chapter. The variational formulation is,
denoting the test function by (y, p) 2 Y ⇥ Y :
(1.136)
Z
Rn
(yy +ry ·ry + py + pp+rp ·rp� yp)dx =
Z
Rn
(fy � pyd)dx.
Set V := H1(Rn
). The Lax-Milgram theorem applies since taking (y, p) = (y, p) weget after cancellation that the bilinear form on the r.h.s. is continuous over V ⇥ Vand satisfies
(1.137) a((y, p), (y, p)) = kyk2V + kpk2V .
7.3. Cheap control.
7. ELLIPTIC CONTROL OVER Rn 29
7.3.1. Zero cost control. Consider a variant of the previous linear quadratic set-
ting, where the control does not appear in the cost function
(1.138) J(u, y) := 12
Z
R(y(x)� yd(x))
2dx,
and there is no control constraints. Then a minimizing sequence could be un-
bounded. Yet we have that, since the problem is convex, a solution (u, y) is charac-terized by the existence of a solution p of the costate equation
(1.139) p��p = y � yd,
with derivative p of the reduced cost equal to 0. This means that y = yd. So, the
problem has a solution i↵ �yd 2 L2(Rn
), i.e. i↵ yd 2 H2(Rn
) and then the unique
solution is u := yd ��yd, y = yd.7.3.2. Small cost control. Assuming again that yd 2 H2
(Rn), let the cost function
be, for " > 0:
(1.140) J"(u, y) :=12"
Z
Rn
u(x)2dx+
12
Z
Rn
(y(x)� yd(x))2dx,
with yd 2 H2(Rn
) and again u := yd ��yd, y = yd. Then J" atttains its minimum
at some (u", y") and
(1.141) ku"k22 +1
"ky" � ydk22
2
"J"(u, y) = kuk22.
So y" ! y in L2(⌦) and u" is bounded, so it has at least one limit-point u. Passing
to the limit in the state equation we obtain that u = y, so that u" * u. By
lemma 1.6, lim sup ku"k2 � kuk2. Combining with the above display we deduce that
ku"k2 ! kuk2. By lemma 1.6 it follows that u" ! u in L2(Rn
).
Remark 1.39. On this ’singular perturbation’ problems (for which the limiting
problem is not of the same nature as the original one), see J.-L. Lions [21].