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Let $I = [0, 1] \subset \mathbb{R}$ and the scalar field is $\mathbb{R}$.

For a Banach space $C(I)$, let $\Lambda(f)=\int_{0}^{1}\left(9 t^{4}-18 t^{3}+11 t^{2}-2 t\right) f(t) d t$

I would like to calculate $\|\Lambda\|$.

I used the fact that f is bounded because it is continuous and defined on a compact set.

Therefore,

$$ |\Lambda f| =\left|\int_{0}^{1}\left(9 t^{4}-18 t^{3}+11 t^{2}-2 t\right) f(t) d t\right| \\ \le \int_{0}^{1}\left|\left(9 t^{4}-18 t^{3}+11 t^{2}-2 t\right) f(t)\right| d t \\ \le \int_{0}^{1}\left|\left(9 t^{4}-18 t^{3}+11 t^{2}-2 t\right)\right| \cdot M d t $$

However, it is impossible to calculate $\int_{0}^{1}\left|\left(9 t^{4}-18 t^{3}+11 t^{2}-2 t\right)\right| d t$, and I got stuck from here.

Would you give some hints about the problem?

alryosha
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    I'm not sure 'impossible' is completely true. Recall the triangle inequality and that you don't need the exact value, an upper bound is enough. – johnny10 May 11 '21 at 09:00

1 Answers1

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Let $p(t)=9 t^{4}-18 t^{3}+11 t^{2}-2 t.$ Then $p$ has the zeros

$$ 0, \frac{1}{3}, \frac{2}{3}, 1.$$

Then we have

$$\int_0^1|p(t)| dt= \int_0^{1/3}(-p(t) dt+\int_{1/3}^{2/3}p(t) dt + \int_{2/3}^1(-p(t) )dt.$$

Fred
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  • Thank you for answering. Can I ask one more question? I obtain $|\Lambda f| \le \frac{49}{810}\cdot M$. Then, How can I obtain $||\Lambda||$? – alryosha May 11 '21 at 09:43
  • @alryosha The Riesz Representation Theorem includes the fact that $||\Lambda||=\int_0^1|9t^4\dots|,dt$. – David C. Ullrich May 11 '21 at 11:13