I am reading Kato's pertubation theory on linear operators and I am not understanding his definition on a quadratic form and polar form. Basically we start with a vector space $V$ and put a symmetric Hermitian inner product $(\cdot,\cdot)$. Now we say that $t : V \times V' -> \mathbb{C}$ is sesquilinear if it is linear on the first and semilinear on the second entry. Notice that the inner product is an example of a sesquilinear form. Now, obviously there's o relation between $t[u,v]$ and $t[v,u]$ for a general form so that he calls $t[u] = t[u,u]$ a quadratic form which needs not be real valued (why, can you give me an example?). He goes on to say that the sesquilinear form $t[u,v]$ is determined by the associated quadratic form $t[u]$. In particular $t[u,v] = 0$ identically if $t[u] = 0$ identically (?). He finishes calling $t[u,v]$ the polar form of $t[u]$.
I kind of see where the quadratic term comes from because if $t = (\cdot,\cdot)$ then $t[u] = t[u,u] = (u,u) = \|u\|^2$ the norm induced by the inner product.
I am not really understanding these definitions; I wonder if anyone could elaborate on them?
Also, there is an exercise: if $|t[u]| \leq M\|u\|^2$ for all vectors $u$ then $|t[u,v]| \leq 2M\|u\|\|v\|.$ I have a strong feeling that we should use the polarization identity for the sesquilinear form $$t[u,v] = \frac{1}{4}(t[u+v]-t[u-v]+it[u+iv]-it[u-iv]),$$and then use the triangle inequality: $$|t[u,v]| \le \frac{1}{4}(|t[u+v]| + |t[u-v]| + |t[u+iv]| + |t[u-iv]|) \leq \frac{M}{4}(\|u+v\|^2 + \|u-v\|^2 + \|u+iv\|^2+\|u-iv\|^2),$$but now I can't put the inner product (to then use the Cauchy-Scharwz) because the $i$'s that multiply $t$ disappeared.
Thank you very much.
Given a linear operator $T:V\to V$, you can define a new sesquilinear form by $$ t[u,v]=(Tu,v). $$ If $T$ is not selfadjoint, the quadratic form will not be real. For example, let $V=\mathbb C^2$, and $T=\begin{bmatrix} 1&2i\\0&0\end{bmatrix}$. This induces $$ t[u,v]:=(Tu,v)=(u_1+2iu_2)\overline{v_1}. $$ This gives you $t[u]=|u_1|^2+2iu_2\overline{u_1}$. Then $t[(1,1)]=1+2i$.
That's by polarization. If $t[u]=0$ for all $u$, then $t[u,v]=0$ for all $u,v$.
Suppose first that $\|u\|=\|v\|=1$. From your inequality, you can use the parallelogram identity. Since $\|iv\|=\|v\|$, \begin{align} \|u+v\|^2 + \|u-v\|^2 + \|u+iv\|^2+\|u-iv\|^2 &=2\|u\|^2+2\|v\|^2+2\|u\|^2+2\|v\|^2\\[0.2cm] &=4\|u\|^2+4\|v\|^2=8. \end{align} In the general case, $$ |t[u,v]|=\|u\|\,\|v\|\,\Big|t\Big[\frac u{\|u\|},\frac v{\|v\|}\Big]\Big| \leq \|u\|\,\|v\|\,\frac M4\,8=2M\|u\|\,\|v\|. $$