Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Problem 16 of chapter 7 states

Use problem 15 to prove that if $u$ is the semigroup solution in $X=L^2(U)$ of $$ \left\{ \begin{array}{rl} u_t - \Delta u =0 & \text{in } U_T \\ u=0 & \text{on } \partial U \times [0,T] \\ u=g & \text{on } U \times \{t=0\}, \end{array} \right. $$ with $g \in C_c^{\infty}(U)$, then $u(\cdot,t) \in C^{\infty}(U)$ for each $0\leq t\leq T$.

Problem 15 states

Let $\{S(t)\}_{t \geq 0}$ be a contraction semigroup on X, with generator $A$. Inductively define $D(A^k):= \{ u \in D(A^{k-1}) \textbf{ | } A^{k-1}u \in D(A)\}$ $(k=2,\dots)$. Show that if $u \in D(A^k)$ for some $k$, then $S(t) u \in D(A^k)$ for each $t \geq 0$.

For problem 15 I didn't need the contraction property for this.

My question is how to use this in the Problem 16. Any help would be appreciated.

share|cite|improve this question

Mmm, you can use 15 to be sure that $S(t)u\in L^2$. Then you may test $$ S(t)u=u+\int_0^t\Delta S(s)u ds $$ against $S(t)u$. Then you observe that $$ \int_0^t\|\nabla S(s)u\|_{L^2}^2ds<\infty $$ Then you have that $$ \|\nabla S(s)u\|_{L^2}^2ds<\infty\,a.e. $$ Pick one of these time points and redo everything. Then you finally get that $$ S(t)u\in H^s,\forall s\geq0 $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.