Best bound for simple inequality

  • Thread starter AiRAVATA
  • Start date
  • #1
173
0
Hello all, the problem I have is the following:

Suppose [itex]f \in C^1(0,1)[/itex] and [itex]f(0) = 0[/itex], then

[tex]
f^2(x) \le \int_0^1 f^2(x) dx,
[/tex]

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

[tex]
f^2(x) \le K \int_0^1 f^2(x) dx,
[/tex]

for K positive?
 

Answers and Replies

  • #2
336
0
Yes,1 is the best possible. Try the functions f_n(x)=x^n.
 
  • #3
173
0
I see. Thanks a lot.
 
  • #4
173
0
Let's see if I got it.

Suppose I want to find the best constant for the inequality

[tex] \int_0^\mu f(x) dx \le K \int_0^1 f(x) dx,[/tex]

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

Let

[tex]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}[/tex]

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

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

so the best constant is [itex]K = 1[/itex].

Is the proof correct?
 
Last edited:

Related Threads on Best bound for simple inequality

  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
5
Views
1K
Replies
2
Views
5K
Replies
1
Views
2K
  • Last Post
Replies
18
Views
5K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
13
Views
11K
  • Last Post
Replies
4
Views
7K
Replies
1
Views
3K
Top