# Best bound for simple inequality

1. Oct 11, 2011

### AiRAVATA

Hello all, the problem I have is the following:

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

$$f^2(x) \le \int_0^1 f^2(x) dx,$$

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

$$f^2(x) \le K \int_0^1 f^2(x) dx,$$

for K positive?

2. Oct 12, 2011

### Eynstone

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

3. Oct 12, 2011

### AiRAVATA

I see. Thanks a lot.

4. Oct 12, 2011

### 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

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

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?

Last edited: Oct 12, 2011