$L^2$ behavior of pseudodifferential operators with symbols in the unit ball

Question

Let $P(x,\xi)$ be the symbol of a pseudodifferential operator. If $|P(x,\xi)|\le 1,\forall\;x,\xi$ and decays off-diagonal, what can we about the behavior of this operator in $L^2$, e.g., is it a contraction mapping?

Answer 1

Your question is rather imprecise. For standard symbols $a(x,\xi)\in S_{1,0}^{m}$ of order $m=0$, we have $L^{2}$-boundedness by the Calderon-Vaillencourt theorem. If you look at the proof via Fourier transforms in Chapter VI of Stein's Harmonic Analysis, you'll see that the operator norm depends on the estimates for derivatives of order $\leq n+1$ (or something like), where $n$ is the dimension of our space. So, accounting for some other implicit dimensional constants, which can easily be made explicit, we pick up along the and which c, if you control the $C_{\alpha\beta}$ in the symbol estimates

$$|\partial_{x}^{\beta}\partial_{\xi}^{\alpha}a(x,\xi)|\leq C_{\alpha\beta}\langle\xi\rangle^{-|\beta|},\quad\forall(x,\xi)\in\mathbb{R}^{n}\times\mathbb{R}^{n}$$

for $|\alpha|,|\beta|\leq n+1$, then you should get that $\|T_{a}\|_{L^{2}\rightarrow L^{2}}<1$.

In general, a symbol in an arbitrary class $S_{\rho,\delta}^{m}$ need not give rise to an $L^{2}$-bounded pseudodifferential operator $T_{a}$. Consider the following example, which I learned from Stein's aforementioned book. Let $\psi\in\mathcal{S}(\mathbb{R})$ such that

$$\widehat{\psi}(\xi)=1, \enspace 2^{-1/4}\leq |\xi| \leq 2^{1/4}; \quad \text{supp}(\psi)\subset\{2^{-1/2}\leq|\xi|\leq 2^{1/2}\}$$

$$a(x,\xi):=\sum_{j=1}^{\infty}e^{-2\pi i 2^{j}x}\widehat{\psi}(2^{-j}\xi),\quad (x,\xi)\in\mathbb{R}\times\mathbb{R}$$

Note that $\widehat{\psi}(2^{-j}\cdot)$ and $\widehat{\psi}(2^{-j'}\cdot)$ have disjoint supports for $j\neq j'$. It's a simple exercise to verify that $a$ satisfies the differential inequalities

$$|\partial_{x}^{\beta}\partial_{\xi}^{\alpha}a(x,\xi)|\leq C_{\alpha\beta}\langle{\xi}\rangle^{-|\alpha|+|\beta|}$$

So $a\in S_{1,1}^{0}$.

Now let $f\neq 0\in\mathcal{S}(\mathbb{R})$ be such that $\text{supp}(\widehat{f})\subset B_{1/2}(0)$. For an integer $N\gg 1$, set

$$f_{N}(x):=\sum_{j=4}^{N}\frac{1}{j}e^{2\pi i 2^{j}x}f(x)$$

By Plancherel,

\begin{align*} \|f_{N}\|_{L^{2}}^{2}=\|\widehat{f_{N}}\|_{L^{2}}^{2}=\int_{\mathbb{R}^{n}}|\sum_{j=4}^{N}j^{-1}\widehat{f}(\xi-2^{j})|^{2}e^{2\pi i x\xi}d\xi=\sum_{j=4}^{N}j^{-2}\|\widehat{f}\|_{L^{2}}^{2}&\leq\left(\sum_{j=4}^{\infty}j^{-2}\right)\|f\|_{L^{2}}^{2} \end{align*}

Since $\psi(2^{-k}\xi)=1$ for $2^{k}-1/2\leq|\xi|\leq 2^{k}+1/2$, $k\geq 4$, we see that

\begin{align*} T_{a}f_{N}(x)&=\sum_{k=4}^{N}\int_{\mathbb{R}}k^{-1}\left(\sum_{j=1}^{\infty}e^{-2\pi i 2^{k}x}\widehat{\psi}(2^{-j}\xi)\right)\widehat{f}(\xi-2^{k})e^{2\pi i x\xi}d\xi\\ &=\sum_{k=4}^{N}k^{-1}\int_{\mathbb{R}}\widehat{f}(\xi-2^{k})e^{2\pi ix(\xi-2^{k})}d\xi\\ &=\sum_{k=4}^{N}k^{-1}f(x)\\ &\sim (\log N)f(x) \end{align*}

Thus, $\|T_{a}f_{N}\|_{L^{2}}\rightarrow\infty$ as $N\rightarrow\infty$, although $\|f_{N}\|_{L^{2}}$ is bounded uniformly in $N$, showing $T_{a}$ is not bounded on $L^{2}$.