c[V]_{A_{r(q)}}^{(r(p)-1)/(r(q)-1)}[U]_{A_{r(q)^{\prime}}}^{[r(p)-r(q)^{\prime}(r(p)-1)]/(r(q)^{\prime}-1)}$
$\displaystyle\simeq
c[V]_{A_{r(q)}}^{(r(p)-1)/(r(q)-1)}[V]_{A_{r(q)}}^{r(p)-r(q)^{\prime}(r(p)-1)}\leq
c_{2}[V]_{A_{r(q)}}^{\alpha(p,q)}.$

Similarly to the above, we obtain

$\left(\int_{{\mathbb{R}}^{n}}|T(x)^{1/p}G(x)|^{p}\,\mathrm{d}x\right)^{q}\leq
c\left(\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}G(x)|^{q}\,\mathrm{d}x\right)^{1/q}.$

Thus, finally, we deduce

$\displaystyle\left(\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}F(x)|^{q}\,\mathrm{d}x\right)^{1/q}\leq
c_{1}K_{p}(c_{2}[V]_{A_{r(p)}}^{\alpha(p,q)})\left(\int_{{\mathbb{R}}^{n}}|W(x)^{1/q}G(x)|^{q}\,\mathrm{d}x\right)^{1/q},$

concluding the proof. ∎

##  8\. Appendix

Here we prove 6.10:

######  6.10.

Set

$C_{n}:=\begin{bmatrix}1&0\\\ 0&n\end{bmatrix},\quad
D_{n}:=\begin{bmatrix}1&\frac{1}{2n^{1/2}}\\\
\frac{1}{2n^{1/2}}&\frac{1}{n}\end{bmatrix},\quad n=1,2,\ldots.$

Then, $C_{n},D_{n}\oldin\nolinebreak[4]\mathrm{P}_{2}({\mathbb{C}})$ with

$\operatorname{tr}(C_{n}D_{n}),\leavevmode\nobreak\
\operatorname{tr}(C_{n}^{-1}D_{n}^{-1})\simeq 1,$

for all $n=1,2,\ldots$. Moreover, for all $a>1$,

$\lim_{n\to\infty}\operatorname{tr}(C_{n}^{a}D_{n}^{a})=\infty.$

Before the proof, we note the identities

(8.1) $\begin{bmatrix}x&z\\\ y&w\end{bmatrix}\begin{bmatrix}\lambda_{1}&0\\\
0&\lambda_{2}\end{bmatrix}\begin{bmatrix}x&y\\\
z&w\end{bmatrix}=\begin{bmatrix}\lambda_{1}x^{2}+\lambda_{2}z^{2}&\lambda_{1}xy+\lambda_{2}zw\\\
\lambda_{1}xy+\lambda_{2}zw&\lambda_{1}y^{2}+\lambda_{2}w^{2}\end{bmatrix}$

and

(8.2) $\operatorname{tr}\left(\begin{bmatrix}1&0\\\
0&\lambda\end{bmatrix}\begin{bmatrix}x&z\\\
y&w\end{bmatrix}\begin{bmatrix}\lambda_{1}&0\\\
0&\lambda_{2}\end{bmatrix}\begin{bmatrix}x&y\\\
z&w\end{bmatrix}\right)=\lambda_{1}x^{2}+\lambda_{2}z^{2}+\lambda\lambda_{1}y^{2}+\lambda\lambda_{2}w^{2}.$

###### Proof (of 6.10).

Let $n$ be any positive integer. It is clear that
$C_{n},D_{n}\oldin\nolinebreak[4]\mathrm{P}_{2}({\mathbb{C}})$. We compute

$C_{n}D_{n}=\begin{bmatrix}1&\frac{1}{2n^{1/2}}\\\
\frac{n^{1/2}}{2}&1\end{bmatrix}$

and

$C_{n}^{-1}D_{n}^{-1}=\begin{bmatrix}1&0\\\
0&\frac{1}{n}\end{bmatrix}\begin{bmatrix}\frac{4}{3}&-\frac{2n^{1/2}}{3}\\\
-\frac{2n^{1/2}}{3}&\frac{4n}{3}\end{bmatrix}=\begin{bmatrix}\frac{4}{3}&-\frac{2n^{1/2}}{3}\\\
-\frac{2}{3n^{1/2}}&\frac{4}{3}\end{bmatrix},$

showing immediately that $\operatorname{tr}(C_{n}D_{n}),\leavevmode\nobreak\
\operatorname{tr}(C_{n}^{-1}D_{n}^{-1})\simeq 1$.

Next, we diagonalize $D_{n}$. Direct computation shows that it has the
eigenvalues

$\lambda_{1}(n):=\frac{1}{2}\left(1+\frac{1}{n}+\sqrt{1-\frac{1}{n}+\frac{1}{n^{2}}}\right)\quad\text{and}\quad\lambda_{2}(n):=\frac{1}{2}\left(1+\frac{1}{n}-\sqrt{1-\frac{1}{n}+\frac{1}{n^{2}}}\right).$

Notice that $\lambda_{1}(n)\simeq 1$ and that

$1-\lambda_{1}(n)=\frac{1}{2}\left(1-\frac{1}{n}-\sqrt{1-\frac{1}{n}+\frac{1}{n^{2}}}\right)=\frac{1}{2}\cdot\frac{-\frac{1}{n}}{1-\frac{1}{n}+\sqrt{1-\frac{1}{n}+\frac{1}{n^{2}}}},$

therefore

$|1-\lambda_{1}(n)|\simeq\frac{1}{n}.$

A normalized eigenvector associated to $\lambda_{1}(n)$ is given by
$e_{1}(n)=(x(n),y(n))$, where

$x(n):=\frac{\frac{1}{2n^{1/2}}}{\sqrt{\frac{1}{4n}+(\lambda_{1}(n)-1)^{2}}},\quad
y(n):=\frac{\lambda_{1}(n)-1}{\sqrt{\frac{1}{4n}+(\lambda_{1}(n)-1)^{2}}}.$

Observe that $|y(n)|\simeq\frac{1}{n^{1/2}}$. Let also $e_{2}(n)=(z(n),w(n))$
be a normalized eigenvector associated to $\lambda_{2}(n)$.

Let now $a>1$ be arbitrary. We have

$D_{n}=\begin{bmatrix}x(n)&z(n)\\\
y(n)&w(n)\end{bmatrix}\begin{bmatrix}\lambda_{1}(n)&0\\\
0&\lambda_{2}(n)\end{bmatrix}\begin{bmatrix}x(n)&y(n)\\\
z(n)&w(n)\end{bmatrix},$

therefore

$\displaystyle D_{n}^{a}$ $\displaystyle=\begin{bmatrix}x(n)&z(n)\\\
y(n)&w(n)\end{bmatrix}\begin{bmatrix}\lambda_{1}(n)^{a}&0\\\
0&\lambda_{2}(n)^{a}\end{bmatrix}\begin{bmatrix}x(n)&y(n)\\\
z(n)&w(n)\end{bmatrix}.$

Thus, by (8.2) it follows that

$\operatorname{tr}(C_{n}^{a}D_{n}^{a})=\lambda_{1}(n)^{a}x(n)^{2}+\lambda_{2}(n)^{a}z(n)^{2}+n^{a}\lambda_{1}(n)^{a}y(n)^{2}+n^{a}\lambda_{2}(n)^{a}w(n)^{2}\gtrsim_{a}n^{a-1}.$

Thus, $\lim_{n\to\infty}\operatorname{tr}(C_{n}^{a}D_{n}^{a})=\infty$. ∎

Identities (8.1) and (8.2) suggest yet another family of counterexamples to
the converse of the Cordes inequality. Let namely

$C_{n}:=\begin{bmatrix}1&0\\\ 0&n(1+\log n)\end{bmatrix}$

and

$D_{n}:=\begin{bmatrix}\sqrt{1-\frac{1}{n}}&-\frac{1}{\sqrt{n}}\\\
\frac{1}{\sqrt{n}}&\sqrt{1-\frac{1}{n}}\end{bmatrix}\begin{bmatrix}1&0\\\
0&\frac{1}{n(1+\log
n)}\end{bmatrix}\begin{bmatrix}\sqrt{1-\frac{1}{n}}&\frac{1}{\sqrt{n}}\\\
-\frac{1}{\sqrt{n}}&\sqrt{1-\frac{1}{n}}\end{bmatrix},$

for all $n=1,2,\ldots$. Then

$(D_{n})_{11}\simeq 1,\quad(D_{n})_{22}\simeq\frac{1}{n},$

therefore

$\operatorname{tr}(C_{n}D_{n})\simeq 1+\log n,$

for all $n=1,2,\ldots$. Thus,
$\lim_{n\to\infty}\operatorname{tr}(C_{n}D_{n})=\infty$.

Let now $a\oldin\nolinebreak[4](0,1)$ be arbitrary. Then, direct computation
gives

$(D_{n}^{a})_{11}\simeq_{a}1,\quad
0\leq(D_{n}^{a})_{12}=(D_{n}^{a})_{21}\lesssim_{a}\frac{1}{\sqrt{n}},\quad(D_{n}^{a})_{22}\simeq_{a}\frac{1}{[n(1+\log
n)]^{a}},$

and

$(D_{n}^{-a})_{11}\simeq_{a}1,\quad|(D_{n}^{-a})_{12}|=|(D_{n}^{-a})_{21}|\lesssim_{a}\frac{[n(1+\log
n)]^{a}}{\sqrt{n}},\quad(D_{n}^{-a})_{22}\simeq_{a}[n(1+\log n)]^{a},$

for all $n=1,2,\ldots$. It follows that for all $m,n=1,2,\ldots$, we have

$\operatorname{tr}(C_{m}^{-a}C_{n}^{a}),\leavevmode\nobreak\
\operatorname{tr}(C_{m}^{-a}D_{n}^{-a}),\leavevmode\nobreak\
\operatorname{tr}(D_{m}^{a}C_{n}^{a})\simeq_{a}\max\left(1,\frac{n(1+\log
n)}{m(1+\log m)}\right)^{a}$

and

$\displaystyle\operatorname{tr}(D_{m}^{a}D_{n}^{-a})$
$\displaystyle\lesssim_{a}\max\left(1,\frac{[n(1+\log
n)]^{a}}{\sqrt{m}\cdot\sqrt{n}},\left(\frac{n(1+\log n)}{m(1+\log
m)}\right)^{a}\right)$
$\displaystyle\lesssim_{a}\max\left(1,\left(\frac{n}{m}\right)^{1/2},\left(\frac{n(1+\log
n)}{m(1+\log m)}\right)^{a}\right).$

Note that an elementary convexity argument gives

$(1-\log 2)(x+y)(1+\log(x+y))\leq x(1+\log x)+y(1+\log y),$

for all $x,y\oldin\nolinebreak[4](0,\infty)$ with $x+y\geq 1$.

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