Does $F(w(x))\leqslant C_wF(x)$ for a certain constant $C_w$?

Question

This question is fully answered, but it is strongly linked to another question I asked later.

While writing a proof, I found myself needing to prove the following result.

Let $F\colon \mathbb Q^n\to \mathbb Q$ be a function satisfying $$F(ty_1,\ldots,ty_n)=\vert t\vert F(y_1,\ldots,y_n)$$ for all $t$ and $(y_1,\ldots,y_n)$. If $w\in \mathrm{GL}_n(\mathbb Q)$, then there exists a constant $C_w$ (which does not depend on $X$), such that $$F(w(X))\leqslant C_wF(X). \quad \forall X \in \mathbb{Q}^n$$

I think I should use something like $\Vert w(W)\Vert\leqslant \Vert w\Vert \cdot \Vert X\Vert$, or try to use a matrix $W=(w_{i,j})$ for $w$ and compute the coordinates of $w(x)$ in terms of $w_{i,j}$ and $x_i$.

But I can't figure it out. Any hint or solution would be much appreciated.

Answer 1

Here's a counterexample : define $F$ as $(x,y)\mapsto \frac{2|xy|}{|x|+|y|}$ if $(x,y)\neq (0,0)$ and $0$ otherwise, and take $w=\left(\begin{smallmatrix}1 & 1 \\ 0 & 1\end{smallmatrix}\right)$. Then $F(w(0,1))=F(1,1)=2$, but $F(0,1)=0$; hence the inequality cannot hold.