# Trace inequality for a trace class operator with powers of $x$ and $\nabla$

I have the following problem. Let $$A$$ be a symmetric positive trace class operator on $$L^2(\mathbb{R})$$ such that $$\mathrm{Tr}(x^6A)$$ and $$\mathrm{Tr}((-\Delta)^3A)$$ are bounded. Is $$\mathrm{Tr}(x^5∇A)$$ bounded as well? Either by algebra formulas or by diagonalizing the operator $$A$$, in which case it a problem looking like Sobolev embeddings with weights ?

(Here, $$x^n$$ is the operator of multiplication by $$x^n$$ and I write $$AB$$ the composition of two operators, so that for example for any $$\varphi\in L^2(\mathbb{R})$$, the operator $$∇A$$ evaluated in $$\varphi$$ yields $$(∇A)\varphi(x) = ∇(A\varphi(x))$$).

Other possible directions:

• A simplified and more analytic version of the same question is the following: if a function $$f∈H^{3}(\mathbb{R},\mathbb{C})$$ and $$\|x^3f\|_{L^2(\mathbb{R},\mathbb{C})} < ∞$$, does it imply that $$∫_{\mathbb{R}} x^5\,\bar{f}\,∇f$$ is bounded?

• Another way to approach the problem: does anyone knows references about non-commutative interpolation theory in weighted spaces?

Remark: Interestingly, it work in the case when $$(6,3,5)$$ is replaced by $$(4,2,3)$$, i.e., trying to bound $$\mathrm{Tr}(x^3∇A)$$ by $$\mathrm{Tr}(x^4A)$$ and $$\mathrm{Tr}(\nabla^4A)$$, since by commutation in the trace and Hölder's inequality for the trace \begin{align*} \left(\mathrm{Tr}(x^3∇A)\right)^2 = \left(\mathrm{Tr}(A^{1/2}x^2x∇A^{1/2})\right)^2 &≤ \mathrm{Tr}(|A^{1/2}x^2|^2)\, \mathrm{Tr}(|x∇A^{1/2 }|^2) \\ &≤ \mathrm{Tr}(x^2Ax^2)\, \mathrm{Tr}(A^\frac{1}{2}∇x^2∇A^\frac{1}{2}) \\ &≤ \mathrm{Tr}(x^4A)\, \mathrm{Tr}(∇x^2∇A) \end{align*} where I use the notation $$|B|^2 := B^*B$$. And using the fact that $$∇x^2 = 2x + x^2∇$$ \begin{align*} \mathrm{Tr}(∇x^2∇A) &= 2\,\mathrm{Tr}(x∇A) + \mathrm{Tr}(x^2∇^2A) \end{align*} and then using again commutation and Hölder's inequality \begin{align*} (\mathrm{Tr}(x∇A))^2 = (\mathrm{Tr}(A^\frac{1}{2}x∇A^\frac{1}{2}))^2 &\leq \mathrm{Tr}(|A^{1/2}x|^2)\, \mathrm{Tr}(|∇A^{1/2 }|^2) \\ &\leq \mathrm{Tr}(x^2A)\,\mathrm{Tr}(\nabla^2A) \\ &\leq \mathrm{Tr}(A)^\frac{1}{2}\mathrm{Tr}(x^4A)^\frac{1}{2}\,\mathrm{Tr}(A)^\frac{1}{2} \mathrm{Tr}(\nabla^4A)^\frac{1}{2} \\ (\mathrm{Tr}(x^2∇^2A))^2 = (\mathrm{Tr}(A^\frac{1}{2}x^2∇^2A^\frac{1}{2}))^2 &\leq \mathrm{Tr}(|A^{1/2}x^2|^2)\, \mathrm{Tr}(|∇^2A^{1/2 }|^2) \\ &\leq \mathrm{Tr}(x^4A)\,\mathrm{Tr}(\nabla^4A) \end{align*}

I will write $$C_0 = \mathrm{Tr}(A)$$, $$X_n = \mathrm{Tr}(x^nA)$$ and $$D_n = \mathrm{Tr}(|\nabla|^6A) = \mathrm{Tr}((-\Delta)^3A)$$. Then the result I get is: $$|\mathrm{Tr}(x^5\nabla A)| ≤ 5 \left(C_0^{1/6}+X_6^{1/12}D_6^{1/12}\right)X_6^{3/4}D_6^{1/12}$$

Remark: For the people interested in functional analysis, it implies \begin{align*} \left|∫x^5\bar{f}\,∇f\,\right| &≤ 5 \left(\|f\|_{L^2}^{1/3} + \||x|^3f\|_{L^2}^{1/6}\|\nabla^3f\|_{L^2}^{1/6}\right) \||x|^3f\|_{L^2}^{3/2}\|\nabla^3f\|_{L^2}^{1/6} \\ &≤ 5\,\|(1+|x|^3)f\|_{L^2}^{5/3}\,\|f\|_{H^3}^{1/3} \end{align*} Remark 2: If $$f$$ is real, we have really better: \begin{align*} \left|∫x^5f\,∇f\,\right| &= \frac{1}{2}\left|∫x^5\,∇(|f|^2)\,\right| = \frac{5}{2}\left|∫x^4\,|f|^2\,\right| \\ &≤ \frac{5}{2}\,\|f\|_{L^2}^{1/3}\,\||x|^3f\|_{L^2}^{2/3} \end{align*}

Proof: Similarly as in the case $$\mathrm{Tr}(x^3\nabla A)$$ I start with \begin{align*} \mathrm{Tr}(x^5\nabla A)^2 &= \mathrm{Tr}(A^{1/2}x^3 x^2\nabla A^{1/2})^2 \\&≤ \mathrm{Tr}(|A^{1/2}x^3|^2)\,\mathrm{Tr}(|x^2∇A^{1/2 }|^2) \\ &\leq X_6\,\mathrm{Tr}(-∇x^4∇A) \end{align*} and I use the fact that $$∇ x^4 = 4 x^3 + x^4\nabla$$ and Hölder's inequality to get \begin{align*} W := \mathrm{Tr}(-∇x^4∇A) &= 4\,\mathrm{Tr}(-x^3∇A) + \mathrm{Tr}(-x^4∇^2A) \\ \\&≤ 4\,\mathrm{Tr}(|A^{1/2}x^3|^2)^{1/2}\,\mathrm{Tr}(|∇A^{1/2 }|^2)^{1/2} + \mathrm{Tr}(|A^{1/2}x^3|^2)^{1/2}\,\mathrm{Tr}(|x∇^2A^{1/2 }|^2)^{1/2} \\ &\leq 4\,X_6^{1/2}\,\mathrm{Tr}(|∇|^2A)^{1/2} + X_6^{1/2}\,\mathrm{Tr}(∇^2x^2\nabla^2A)^{1/2} \end{align*}

The first term is bounded by interpolation by $$\mathrm{Tr}(|∇|^2A)≤ C_0^{2/3}D_6^{1/3}$$. For the second I use $$∇x^2 = 2x+x^2\nabla$$, and again Hölder's inequality and the commutation in the trace to get \begin{align} \mathrm{Tr}(∇^2x^2\nabla^2A) &= 2\,\mathrm{Tr}(∇x\nabla^2A) + \,\mathrm{Tr}(\nabla x^2\nabla^3A) \\ &≤ 2 \,\mathrm{Tr}(-∇x^2\nabla A)^{1/2}\,\mathrm{Tr}(∇^4 A)^{1/2} + \mathrm{Tr}(-∇x^4\nabla A)^{1/2}\,D_6^{1/2} \end{align}

• To bound the first term I do as in the case $$\mathrm{Tr}(x^3\nabla A)$$ (treated in my question) to get $$\mathrm{Tr}(-∇x^2\nabla A)≤ 2\,C_0^{1/2}X_4^{1/4}D_4^{1/4}+X_4^{1/2}D_4^{1/2}$$ and then by interpolation $$D_4≤ C_0^{1/3}D_6^{2/3}$$, $$X_4≤ C_0^{1/3}X_6^{2/3}$$ so that \begin{align*} \mathrm{Tr}(-∇x^2\nabla A)\,\mathrm{Tr}(∇^4 A) &≤ (2\,C_0^{1/2}X_4^{1/4}D_4^{1/4}+X_4^{1/2}D_4^{1/2})D_4 \\ &≤ 2\,C_0^{1/2}X_4^{1/4}D_4^{5/4}+X_4^{1/2}D_4^{3/2} \\ &≤ 2\,C_0\,X_6^{1/6}D_6^{5/6}+C_0^{2/3}X_6^{1/3}D_6 =: C^2 \end{align*}

• Now I remark that the second term is nothing but $$W^{1/2}D_6^{1/2}$$, that I was trying to control, and so putting things together and using the fact that by Young's inequality $$ab≤a^4/4+3b^{4/3}/4$$, \begin{align*} W &≤ 4 C_0^{1/3}X_6^{1/2}D_6^{1/6}+X_6^{1/2}(2C+W^{1/2}D_6^{1/2})^{1/2} \\ &≤ 4 C_0^{1/3}X_6^{1/2}D_6^{1/6}+\sqrt{2}\,C^{1/2}X_6^{1/2}+X_6^{1/2}W^{1/4}D_6^{1/4} \\ &≤ 4 C_0^{1/3}X_6^{1/2}D_6^{1/6}+\sqrt{2}\,C^{1/2}X_6^{1/2}+\frac{3}{4}\,X_6^{2/3}D_6^{1/3} + \frac{1}{4} W \end{align*} so that putting all the $$W$$ on the same side \begin{align*} W &≤ \frac{16}{3} C_0^{1/3}X_6^{1/2}D_6^{1/6}+\frac{4\sqrt{2}}{3}\,C^{1/2}X_6^{1/2}+X_6^{2/3}D_6^{1/3} \end{align*} and so $$\mathrm{Tr}(x^5\nabla A) ≤ X_6^{1/2}W^{1/2} ≤ \left(\frac{16}{3} C_0^{1/3}X_6^{3/2}D_6^{1/6}+\frac{4\sqrt{2}}{3}\,\left(2\,C_0\,X_6^{1/6}D_6^{5/6}+C_0^{2/3}X_6^{1/3}D_6\right)^{1/4}X_6^{3/2}+X_6^{5/3}D_6^{1/3}\right)^{1/2}$$ We can use Young's inequality to simplify and get something of the form with $$C≤5$$ $$\mathrm{Tr}(x^5\nabla A) ≤ C \left(C_0^{1/6}+X_6^{1/12}D_6^{1/12}\right)X_6^{3/4}D_6^{1/12}$$