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I have a question regarding a PDE course:

Let $T$ be the strongly continuous semigroup which belongs to the heat equation, thus with generator $A$ is the Laplacian.

Suppose we have $g \in C^{\infty}_c(U)$, and define the sets \begin{align*} D(A) & = H^1_0(U) \cap H^2(U) \\ D(A^2) & = \{ x \in D(A) \mid Ax \in D(A)\} \\ & = \{x \in H^1_0(U) \cap H^2(U) \mid \Delta x \in H^1_0(U) \cap H^2(U) \} \\ D(A^k) & = \{ x \in D(A^{k-1}) \mid \Delta x \in H^1_0(U) \cap H^2(U) \}. \end{align*}

We now want to prove that $$T(t)g \in C^{\infty}(U).$$

First I prove that $g \in D(A^k)$ for all $k$, using induction. For $k=0$ we have $$g \in C^{\infty}_C(U) \subseteq \overline{C^{\infty}_C(U)} = H^1_0(U).$$ We know that $g$ and all the derivatives of $g$ are compactly supported, thus we have $$\int_U \lvert g \rvert^2 \,\mathrm{d}x \leq \lvert U\rvert \sup_u g^2 < \infty,$$ where the last inequality follows from the fact that $U$ is bounded. For the derivatives of $g$ we have a similar proof, thus $g \in H^2(U)$, therefore $g \in D(A)$.

Now suppose it is true for $k-1$, we now want to prove it for $k$. Suppose $g \in D(A^{k-1})$, then $A^{k-1}g = \Delta^{k-1}g \in C^{\infty}_C(U)$. In a similar way as above I can prove that $g \in D(A^k)$.

I have also proven that from this it follows that $T(t)g \in D(A^k)$ for all $k$. Then $$T(t)g \in \cap_{k=1}^{\infty} D(A^k),$$ however this is where I cannot continue my proof. The last expression should be embedded in $C^{\infty}(U)$.

Can someone give me a hint or a solution.

If I have proven that, I can complete the rest of the proof myself, however, I am stuck on this.

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  • $\begingroup$ What can you say about the Fourier tranform of $h \in \bigcap_{k=1}^{\infty}D(A^{k})$ by starting with $C_{c}^{\infty}$? $\endgroup$ Jun 4, 2015 at 23:34
  • $\begingroup$ I don't understand what you mean. $\endgroup$
    – nippon
    Jun 8, 2015 at 13:09
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    $\begingroup$ Start with $\bigcap_{k=1}^{\infty}D(A^{k}) \subseteq \bigcap_{k=1}^{\infty}H^{2k}(U)$. $\endgroup$ Jun 8, 2015 at 15:04
  • $\begingroup$ But won't this all show that the final solution is weakly differentiable? Do we use Fourier to show that it is even regularly (= non-weakly) differentiable? $\endgroup$ Jun 8, 2015 at 15:18
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    $\begingroup$ Look at Morrey's embedding, it says that you can trade weak derivatives for classical derivatives at a certain cost. $\endgroup$
    – Jose27
    Jan 1, 2018 at 18:59

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