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In page 79 of Davies's book on Heat Kernels and spectral theory, the author proves that $$\lVert e^{-Ht}f \rVert_2 \leq c_1t^{-\mu/ 4}\lVert f \rVert_1$$ where the norms are $L^p$ norms. He states

by duality, it follows that $$\lVert e^{-Ht}f \rVert_\infty \leq c_1t^{-\mu/ 4}\lVert f \rVert_2$$

Can someone explain what exactly this "duality" argument is? Above, $e^{-Ht}f$ we can take to be the solution of the heat equation with initial data $f$ (where $H$ is the Laplacian).

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    $\begingroup$ I do not think this is research level. However, I do think it would be useful to record a discussion of certain duality arguments, maybe over at StackExchange. See Q1 Q2 Q3 $\endgroup$
    – user70229
    Commented Feb 9, 2016 at 14:10

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You have shown that $e^{-Ht}\colon L^1\to L^2$ is bounded with the given operator norm. It follows that the dual operator $$\left(e^{-Ht}\right)^*\colon (L^2)^* \to (L^1)^*$$ is bounded with the same operator norm, because of the characterisation $$\|T\|_{L^1\rightarrow L^2}=\|T^*\|_{L^\infty\rightarrow L^2}=\sup_{f,g}\frac{\langle Tf,g\rangle}{\|f\|_1\|g\|_2}\,.$$
Since $(L^1)^* = L^\infty$ and $(L^2)^* = L^2$ (these are isomorphisms, actually), it remains to check that $\left(e^{-Ht}\right)^* = e^{-Ht}$.

For this note that $$\int f(x)(e^{-Ht}g)(x)dx = \iint f(x)e(x,y;t)g(y)dxdy$$ where $e(x,y;t)$ is the heat kernel, which is real and symmetric in the variables $x,y$. It follows that $\left(e^{-Ht}\right)^*$ is again integration against the heat kernel as claimed.

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    $\begingroup$ Thanks. your domain and range are mixed in the first line btw. $\endgroup$
    – FavorExistingPopularTags
    Commented Feb 9, 2016 at 13:32
  • $\begingroup$ In the second line too. Fixed now. $\endgroup$
    – Lior Silberman
    Commented Feb 9, 2016 at 13:48
  • $\begingroup$ In the last step, all we really need is that the semigroup $e^{-Ht}$ is a bounded self-adjoint operator, which follows from the fact that $H$ is self-adjoint. We don't need to use the fact that a heat kernel exists, which is quite a bit more difficult to prove. $\endgroup$
    – Nate Eldredge
    Commented Feb 9, 2016 at 15:09
  • $\begingroup$ This is not obvious to me. I agree that $T=\exp(-Ht)\colon L^2\to L^2$ has $T^*=T$, but I don't see why that automatically shows something about the dual of $S=\exp(-Ht)\colon L^1\to L^2$. $\endgroup$
    – Lior Silberman
    Commented Feb 9, 2016 at 15:32
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    $\begingroup$ Basically you differentiate both sides wrt $t$ and use the self-adjointness of the Laplace operator, but this brushes asides various minor technical issues. $\endgroup$
    – Lior Silberman
    Commented Feb 9, 2016 at 17:29

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