# 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!

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.

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.