What it means if a functional is bounded

functional-analysissesquilinear-forms

I'm learning the definition of quadratic and sesquilinear forms. I found two formulations of what it means if a sesquilinear form $\tilde q$ is bounded.

  1. $|\tilde q(\phi,\psi)|\leq c||\phi||\ ||\psi||$ with $c<\infty$.

and

  1. $\|\tilde q\|=\sup_{\|\phi\|=\|\psi\|=1}|\tilde q(\phi,\psi)|$ is finite.

I can see both definitions included what it means to be bounded, but I'm a bit confused with the notation. Is there any difference between $|\tilde q(\phi,\psi)|$ and $\|\tilde q\|$? Do they both refer to the norm of a sesquilinear form (I actually don't know what that means)? Also for the first definition, how does the inequality come from?

Thanks for the help!

Best Answer

If $f: \mathcal H \times \mathcal H \to \mathbb C$ is a sesquilinear form on a Hilbert space $\mathcal H$, we define its norm as $$\|f\|:= \sup\{|f(x,y)|: x,y\in \mathcal H, \|x\| = \|y\| = 1\}$$ If $\|f\| < \infty$, we say that the sesquilinear form $f$ is bounded.

That resolves your questions about (2). See the definition here also.


I shall prove (1), i.e., $$|f(x,y)| \le c \|x\| \|y\|$$ for all $x,y\in \mathcal H$. Moreover, we shall show that $c = \|f\|$ as defined above.

If either of $x$ or $y$ are $0$, the inequality is trivial. So, assume $x,y\ne 0$ and $x,y\in \mathcal H$. Consider $\frac{x}{\|x\|}$ and $\frac{y}{\|y\|}$. We have $$\frac{f(x,y)}{\|x\|\|y\|} = f\left(\frac{x}{\|x\|}, \frac{y}{\|y\|} \right)$$ and $$\left|f\left(\frac{x}{\|x\|}, \frac{y}{\|y\|} \right) \right| \le \|f\|$$

since $\frac{x}{\|x\|}$ and $\frac{y}{\|y\|}$ have unit norm. It follows that $$|f(x,y)| \le \|f\| \|x\|\|y\|$$ for all $x,y\in\mathcal H$.


Remark. Since $c = \|f\|$ (as we have shown above), it is not too hard to see that the two notions of boundedness as you describe, are equivalent.

I hope that helps. Let me know if you have any questions.