Best bound for simple inequality

[AiRAVATA]

Hello all, the problem I have is the following:

Suppose f \in C^1(0,1) and f(0) = 0, then

<br /> f^2(x) \le \int_0^1 f^2(x) dx,<br />

but I was wondering if 1 is the best constant for the inequality. In other words, how do I determine the best bound for

<br /> f^2(x) \le K \int_0^1 f^2(x) dx,<br />

for K positive?

 

[Eynstone]

Yes,1 is the best possible. Try the functions f_n(x)=x^n.

 

[AiRAVATA]

I see. Thanks a lot.

 

[AiRAVATA]

Let's see if I got it.

Suppose I want to find the best constant for the inequality

\int_0^\mu f(x) dx \le K \int_0^1 f(x) dx,

where f(x) \in C^1(0,1), f(0) = 0, f(x) \ge 0, and 0 \le \mu \le 1.

Let

f_n(x) = \begin{cases} \frac{n+2}{n+3} x(2\mu -x), &amp;0 &lt; x \le \mu, \\<br /> \\<br /> \frac{n+2}{n+3} \mu^2 \mbox{sech} [(n+2)(\mu-x)], &amp;\mu &lt; x &lt; 1. \end{cases}

If s_n = \int_0^\mu f(x) dx, and S_n = \int_0^1 f(x) dx, then

\bigl\{s_n\bigr\} \nearrow \frac{2 \mu^3}{3}, \mbox{ and }\bigl\{S_n\bigr\} \searrow \frac{2 \mu^3}{3},

so the best constant is K = 1.

Is the proof correct?