I have the following problem (which was posed an exercise): Let E be a real vector space with a symplectic form $\omega$ (that is a non-degenerate skew-symmetric bilineal form).
Suppose you have another skew symmetric form $\tau$ so that for some $C>0$:
$$ | \tau(x,y) | \leq C| \omega(x,y) | , \forall x,y\in E $$
Then there exists $c\in\mathbb{R}$ so that $\tau = c\omega$. I solved the problem like this:
Pick a symplectic basis of E: $\{x_1,\ldots,x_n,y_1,\ldots,y_n\}$, that is a basis so that $ 0=\omega(x_i,x_j)=\omega(y_i,y_j) $, $ \omega(x_i,y_j)= \delta_{i,j} $. This basis always exists when there is a symplectic form.
Then the equation above implies that $\tau$ only depends on $\tau(x_i,y_i)=c_i$. (Since al the other values of $\tau$ on basic elements are $0$). If $c_i\neq c_j$ for some pair indexes, then: $$ 0<|c_1-c_2| = |\tau(x_i,y_i)-\tau(x_j,y_j)| = |\tau(x_i,y_i-y_j)-\tau(x_j,y_j-y_i) | = |\tau(x_1+x_2,y_1-y_2)| \leq |\omega(x_1+x_2,y_1-y_2)| = |\omega(x_i,y_i)-\omega(x_j,y_j)|=0 $$
Which obviously is a contradiction. Is my proof right?
The reason I am suspicious is because the excercise in question ask you first about the case $\tau$ is skew-symmetric and then when it is not. Which gives me the impression that it either the result should be false when $\tau$ is not skew-symmetric or this case should be harder. But in my solution, which seems quite easy it not necessary to assume $\tau$ has any symetry.