Norm of a linear continuous form

Question

Let $E=\{f\colon[0,2]\to\mathbb{R} \mid f \text{ continuous} \}$ be a prehilbert space equipped with inner product: $$\langle f,g\rangle=\int_0^2 f(t)g(t)\, dt$$ And let : $$U\colon E \to\mathbb{R}$$ $$f \mapsto \int_0^1 t^2 f(t)\, dt$$ I have proved that $u$ is linear and continuous, now show that : $$\frac{1}{2\sqrt{3}}\leq\|u\|\leq \frac{1}{\sqrt{5}}$$ Hint:take $f(t)=\frac{1}{\sqrt{2}}$

I have proved $\|u\|\leq \frac{1}{\sqrt{5}}$, I need the first part of the inequality

I think you mean $1/(2 \sqrt{3})$, not $(1/2) \sqrt{3}$. – Robert Israel May 14 '15 at 15:34 — Robert Israel, May 14 '15 at 15:34

score 0 · Accepted Answer · answered May 14 '15 at 15:36

0

To get a lower bound on $\|U\|$, take some nice function $f$ and compute $|U(f)|/\|f\|$. But you can do better than $1/(2\sqrt{3})$. You might as well have $f = 0$ on $[1,2]$ since that interval doesn't contribute to $U(f)$.

answered May 14 '15 at 15:36

Robert Israel

448,999

For f(t)=1/sqrt(2), u(t)=1/(2 sqrt(3)) – Math1995 May 14 '15 at 15:41
It's not clear enough to me – Math1995 May 14 '15 at 15:45

Norm of a linear continuous form

1 Answers1