Lecture 1. Semigroups in Banach spaces: The Hille-Yosida theorem

Posted on January 14, 2019 by Fabrice Baudoin

The first few lectures will be devoted to some elements of  the general theory of operators in Banach and Hilbert spaces which are useful when studying Dirichlet forms. In this lecture, we focus on the Hille–Yosida theorem.

Let (B,\| \cdot \|) be a Banach space, which for us will later be L^p(X,\mu), where (X,\mu) is a measure space.

Definition:  A family of bounded operators (P_t)_{t \ge 0} on B is called a strongly continuous contraction semigroup if:

  1. T_0 =\mathbf{Id} and for s,t \ge 0, P_{s+t}=P_s P_t;
  2. For every  x \in B, the map  t \to P_t x is continuous;
  3. For every  x \in B and t \ge 0, \| P_t x \| \le \|x \|.

We  recall that a densely defined linear operator A: \mathcal{D}(A) \subset B \to B is said to be closed if for every sequence x_n \in \mathcal{D}(A) that converges to x \in B and such that Ax_n \to y \in B, we have x \in \mathcal{D}(A) and y=Ax.

Proposition: Let (P_t)_{t \ge 0} be a strongly continuous contraction semigroup on B. There exists a closed and densely defined operator A: \mathcal{D}(A) \subset B \to B where

\mathcal{D}(A)=\left\{ x \in B,\quad \lim_{t \to 0} \frac{P_t x -x}{t} \text{ exists} \right\},
such that for x \in \mathcal{D}(A),

\lim_{t \to 0} \left\| \frac{P_t x -x}{t} -Ax \right\|=0.
The operator A is called the generator of the semigroup (P_t)_{t \ge 0}. We also say that A generates (P_t)_{t \ge 0}.

Proof: Let us consider the following bounded operators on B :

A_t= \frac{1}{t} \int_0^t P_s ds.

For f \in B and h>0, we have
\frac{1}{t} \left( P_t A_h f -A_h f \right)  = \frac{1}{ht} \int_0^h (P_{s+t}f -P_{s}f )ds
= \frac{1}{ht} \int_0^t (P_{s+h}f -P_{s}f )ds.

Therefore, we obtain
\lim_{t \to 0} \frac{1}{t} \left( P_t A_h f -A_h f \right)=\frac{1}{h} \left( P_h f - f \right).
This implies,

\left\{ A_h f, f \in B, h >0 \right\} \subset \left\{ f \in B,\quad \lim_{t \to 0} \frac{P_t f -f}{t} \text{ exists} \right\}
Since \lim_{h \to 0} A_h f =f, we deduce that

\left\{ f \in B,\quad \lim_{t \to 0} \frac{P_t f -f}{t} \text{ exists} \right\}
is dense in B. We can then consider

Af = \lim_{t \to 0} \frac{P_t f -f}{t} ,
which is of course defined on the domain
\mathcal{D}(A)=\left\{ f \in B,\quad \lim_{t \to 0} \frac{P_t f -f}{t} \text{ exists} \right\}. \square

We may observe that the proof of the above result does not involve the contraction property of (T_t)_{t \ge 0}, so that it may be extended to strongly continuous semigroups.

Let A: \mathcal{D}(A) \subset B \rightarrow B be a densely defined closed operator. A constant \lambda \in \mathbb{R} is said to be in the spectrum of A if the operator \lambda \mathbf{Id}-A is not bijective. In that case, it is a consequence of the closed graph theorem that if \lambda is not in the spectrum of A , then the operator \lambda \mathbf{Id}-A has a bounded inverse. The spectrum of an operator A shall be denoted \rho(A).

The following important theorem is due to Hille and Yosida and provides, through spectral properties, a characterization of closed operators that are generators of contraction semigroups.

Theorem: (Hille-Yosida theorem) A necessary and sufficient condition that a densely defined closed operator A generates a strongly continuous contraction semigroup is that:

  1. \rho (A) \subset (-\infty,0];
  2. \| (\lambda \mathbf{Id} -A)^{-1} \| \le \frac{1}{\lambda} for all \lambda > 0.

Proof: Let us first assume that A generates a strongly continuous contraction semigroup (T_t)_{t \ge 0}. Let \lambda >0 . We want to prove that \lambda \mathbf{Id}-A is a bijective operator \mathcal{D}(A) \rightarrow B. The formal Laplace transform formula
\int_0^{+\infty} e^{-\lambda t} e^{tA} dt=(\lambda \mathbf{Id}-A)^{-1},
suggests that the operator
\mathbf{R}_\lambda =\int_0^{+\infty} e^{-\lambda t} T_t dt
is the inverse of \lambda \mathbf{Id}-A.  We prove this is the case. First, let us observe that \mathbf{R}_\lambda is well-defined as a Riemann integral since t \to T_t is continuous and \| T_t \| \le 1.  We now show that for x \in B, \mathbf{R}_\lambda x \in \mathcal{D}(A). For h > 0, we have
\frac{T_h -\mathbf{Id}}{h} \mathbf{R}_\lambda x = \int_0^{+\infty} e^{-\lambda t} \frac{T_h -\mathbf{Id}}{h} T_t x dt
=\int_0^{+\infty} e^{-\lambda t} \frac{T_{h+t} -T_t}{h} x dt
= e^{\lambda h} \int_h^{+\infty} e^{-\lambda s} \frac{T_{s} -T_{s-h}}{h} x ds
=\frac{e^{\lambda h}}{h} \left( \mathbf{R}_\lambda x -\int_0^h e^{-\lambda s} T_{s} x ds- \int_h^{+\infty} e^{-\lambda s} T_{s-h} xds \right)
=\frac{e^{\lambda h}-1}{h} \mathbf{R}_\lambda x-\frac{e^{\lambda h}}{h}\int_0^h e^{-\lambda s} T_{s} x ds

By letting h \to 0, we deduce that \mathbf{R}_\lambda x \in \mathcal{D}(A) and moreover A \mathbf{R}_\lambda x =\lambda \mathbf{R}_\lambda x -x.
Furthermore, it is readily checked that, since A is closed, for x \in \mathcal{D}(A),

A \mathbf{R}_\lambda x=A\int_0^{+\infty} e^{-\lambda t} T_t xdt= \int_0^{+\infty} e^{-\lambda t} A T_t xdt

=\int_0^{+\infty} e^{-\lambda t} T_t A xdt =\mathbf{R}_\lambda A x.
We therefore conclude
(\lambda \mathbf{Id}-A) \mathbf{R}_\lambda =\mathbf{R}_\lambda (\lambda \mathbf{Id}-A)=\mathbf{Id}.
Thus,
\mathbf{R}_\lambda=(\lambda \mathbf{Id}-A)^{-1},
and it is clear that
\| \mathbf{R}_\lambda \| \le \frac{1}{\lambda}.

Let us now assume that A is a densely defined closed operator that satisfies the two assumptions of the theorem.

The idea is to consider the following sequence of bounded operators
A_n=-n \mathbf{Id}+n^2 (n\mathbf{Id}-A)^{-1},
from which it is easy to define a contraction semigroup and then to show that A_n \to A. We will then define a contraction semigroup associated to A as the limit of the contraction semigroups associated to A_n.

First, for x \in \mathcal{D}(A), we have
A_n x =n(n\mathbf{Id}-A)^{-1} A x \to_{n \to +\infty} 0.
Now, since A_n is a bounded operator, we may define a semigroup (T_t^n)_{t \ge 0} through the formula
T_t^n =\sum_{k=0}^{+\infty}\frac{t^k A_n^k}{k !}.
At that point, let us observe that we also have
T_t^n =e^{-nt} \sum_{k=0}^{+\infty}\frac{n^{2k} t^k (n\mathbf{Id}-A)^{-k}}{k !}.
As a consequence, we have
\|T_t^n \|  \le e^{-nt} \sum_{k=0}^{+\infty}\frac{n^{2k} \| (n\mathbf{Id}-A)^{-1}\|^k }{k !}

\le e^{-nt} \sum_{k=0}^{+\infty}\frac{n^{k} t^k }{k !}  \le 1
and (T_t^n)_{t \ge 0} is therefore a contraction semigroup. The strong continuity is also easily checked:
\| T_{t+h}^n -T_t^n \|  = \| T_t^n (T_{h}^n -\mathbf{Id})\|
 \le \| T_{h}^n -\mathbf{Id})\|
\le \sum_{k=1}^{+\infty}\frac{h^k \|A_n\|^k}{k !} \to_{h \to 0} 0.

We now prove that for fixed t \ge 0, x \in \mathcal{D}(A), (T_t^n x)_{n \ge 1} is a Cauchy sequence. We have
\| T_t^n x -T_t^m x \|  =\left\| \int_0^t \frac{d}{ds} (T_s^n T_{t-s}^m x) ds \right\|
 = \left\| \int_0^t T_s^n T_{t-s}^m (A_n x-A_mx) ds \right\|
 \le \int_0^t \| A_n x-A_mx \| ds

 \le t \| A_n x-A_mx \|.

Therefore for x \in \mathcal{D}(A), (T_t^n x)_{n \ge 1} is a Cauchy sequence and we can define

T_t x =\lim_{n \to +\infty} T_t^n x.
Since \mathcal{D}(A) is dense and the family (T_t^n )_{n \ge 1} uniformly bounded, the above limit actually exists for every x \in B, so that (T_t)_{t \ge 0} is well-defined on B. It is clear that (T_t)_{t \ge 0} is a strongly continuous semigroup, inheriting these properties from (T_t^n)_{t \ge 0} (the details are let to the reader here).

It remains to show that the generator of (T_t)_{t \ge 0}, call it \tilde{A} is equal to A. For every t \ge 0, x \in \mathcal{D}(A) and n \ge 1,

T_t^n x =x +\int_0^t T_s^n Ax ds,
therefore

T_t^n x =x +\int_0^t T_s^n Ax ds.
Hence \mathcal{D}(A) \subset \mathcal{D}(\tilde{A}) and for x \in \mathcal{D}(A), \tilde{A} x=Ax. Finally, since for \lambda > 0, (\lambda \mathbf{Id}-A) \mathcal{D}(A)=B=(\lambda \mathbf{Id}-\tilde{A}) \mathcal{D}(\tilde{A}), we conclude \mathcal{D}(A) = \mathcal{D}(\tilde{A}). \square

Exercise:  By using the proof of Hille-Yosida theorem, show the following fact: If A_1 and A_2 are the generators of contraction semigroups (T_t^1)_{t \ge 0} and (T_t^2)_{t \ge 0}, then for x \in B, the two following statements are equivalent:

  1.  \forall \text{ } \lambda > 0, \quad (\lambda \mathbf{Id}-A_1)^{-1}x= (\lambda \mathbf{Id}-A_2)^{-1}x;
  2. \forall \text{ } t \ge 0, \quad T_t^1x=T_t^2 x.

 

As powerful as it is, the Hille-Yosida theorem is difficult to directly apply to the theory of diffusion semigroups. The result of the following exercise is useful in concrete situations.
Exercise:   A densely defined operator on a Banach space B is called dissipative if for each x \in \mathcal{D}(A), we can find an element \phi of the dual space B^*, such that:

  1.  \| \phi \|=\|x \|;
  2.  \phi (x)=\| x\|^2;
  3. \phi (A x) \le 0.

Show that a closed operator A on a Banach space B is the generator of a strongly continuous contraction semigroup, if and only if:

  1.  A is dissipative;
  2.  For \lambda > 0, the range of the operator \lambda \mathbf{Id} -A is B.
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