first off this is my first time posting here so I am only learning how to format questions. Please bear with me.
I am trying to prove the asymptotic expansion for the symbol of the composition of two pseudodifferential operators (of class $S_{1,0}$). The expansion itself I have found but I am having trouble with the remainder term. Supposing $p(x,\xi)$ and $q(x,\xi)$ are symbols of order $m_1$ and $m_2$ respectively (meaning $|D_x^\beta D_\xi^\alpha \,p(x,\xi)|\le C_{\alpha,\beta}\,(1+|\xi|)^{m_1-|\alpha|}$ and similarly for $q(x,\xi)$). I need to show $$|D_x^\beta D_\xi^\alpha \,R_N(x,\xi)|\le C_{\alpha,\beta}\,(1+|\xi|)^{m_1+m_2-|\alpha|-N}$$ for any $N\gt 0$. This remainder term is given by $$R_N(x,\xi)=\int r_N(x,\xi,\eta) \,\widehat{q(\eta,\xi)}\,e^{ix\eta}\,d\eta $$ where $r_N(x,\xi,\eta)$ is the integral form of the remainder from Taylor's Theorem: $$r_N(x,\xi,\eta)=\sum_{|\delta|=N}\frac{N}{\delta !}\; \eta^\delta \;\int_0^1(1-t)^N\,D_\xi^\delta \,p(x,\xi+t\eta)\,dt $$
Applying Leibniz rule we get $$|D_x^\beta D_\xi^\alpha\,R_N(x,\xi)|\le\int|D_x^\beta D_\xi^\alpha [r_N(x,\xi,\eta)\, \widehat{q(\eta,\xi)}\,e^{ix\eta}\,]|d\eta $$ $$\le \int \sum_{\lambda \le \alpha} \sum_{\mu \le \beta}|D_x^\mu D_\xi^\lambda\,r_n(x,\xi,\eta)|\;|D_\xi^{\alpha-\lambda}\,\widehat{q(\eta,\xi)}|\;|D_x^{\beta-\mu}\,e^{ix\eta}|\,d\eta $$
Assuming $q(x,\xi)$ has compact support in $x$, using integration by parts we get the estimate $$|D_\xi^{\alpha-\lambda}\,\widehat{q(\eta,\xi)}|\le C_{M,\alpha,\lambda} \frac{(1+|\xi|)^{m_2-|\alpha-\lambda|}}{(1+|\eta|)^M}$$ for any $M\ge 0$. So what is left is estimates for $r_n(x,\xi,\eta)$. We get $$ |D_x^\mu D_\xi^\lambda\,r_n(x,\xi,\eta)|\le C_N \; |\eta|^N \sum_{|\delta|=N} max_{\,t\in [0,1]} |D_x^\mu D_\xi^{\lambda+\delta}\,p(x,\xi+t\eta)| $$ $$ \le C_{N,\mu,\delta,\lambda} \; |\eta|^N \sum_{|\delta|=N} max_{\,t\in [0,1]} (1+|\xi+t\eta|)^{m_1-|\lambda+\delta|}$$
From here the idea is to split the integral in two parts for $|\eta|\le |\xi|/2$ and the rest. The rest is fine and I get the desired bound for that part taking $M$ large enough. The trouble is with the first part. Using that $|\eta|\le |\xi|/2$ then we can get the bound $$ |D_x^\mu D_\xi^\lambda\,r_n(x,\xi,\eta)|\le C_{N,\mu,\delta,\lambda} \; |\eta|^N \sum_{|\delta|=N}(1+|\xi|)^{m_1-|\lambda+\delta|}$$
Applying all of this gives (ignoring the other part of the integral which is well bounded) $$|D_x^\beta D_\xi^\alpha \,R_N(x,\xi)|\le C_{} \sum_{\lambda \le \alpha} \sum_{|\delta|=N}(1+|\xi|)^{m_1+m_2-|\alpha-\lambda|-|\lambda+\delta|}\,\int_{|\eta| \le |\xi|/2} \frac {|\eta|^{\beta+N}}{(1+|\eta|)^M} d\eta$$
This integral is obviously converges and using triangle inequality in the exponent of $(1+|\xi|)$ gives
$$|D_x^\beta D_\xi^\alpha \,R_N(x,\xi)|\le C_{\alpha,\beta} \sum_{|\delta|=N}(1+|\xi|)^{m_1+m_2-|\alpha+\delta|}$$
So finally (this seems long winded) I have arrived at my problem. Since $-|\alpha+\delta| \ge -|\alpha| - |\delta| = -|\alpha| - N$, I don't end up getting my desired bound.
I wrote a lot down, possibly unnecessarily but I am going mad thinking it is something very obvious I am missing. Can anyone spot my mistake?? Also please let me know if my formatting needs work.
Cheers!
In fact I have found my error and it was very embarrassing, simply a matter of notation.
Of course $\alpha$ and $\delta$ are multi indices so the norm on them is not the euclidean norm. For multi indices we have obviously $|\alpha + \delta|= |\alpha|+ |\delta|$ and so everything is just as it should be, triangle inequality does not come into play. I get my desired bound.
A brain fart cost me a few days.