Is the 'Integer Function' Bound Valid for Given Functions f(t) and g(u)?

  • Context: Graduate 
  • Thread starter Thread starter zetafunction
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary
SUMMARY

The discussion centers on the validity of the 'Integer Function' bound for the integral expression involving functions f(t) and g(u). The proposed bound is expressed as \int_{0}^{\infty}dt [g(x/t)]f(t) \le \int_{0}^{\infty}dt g(x/t)f(t), where g is a non-decreasing positive function. The participants suggest that incorporating g(x/t) + 1 for f+(t) and g(x/t) - 1 for f-(t) may yield a more effective bound. This approach leverages the properties of f(t) being non-negative and the behavior of g(u) under the specified conditions.

PREREQUISITES
  • Understanding of integral calculus and bounds
  • Familiarity with properties of non-decreasing functions
  • Knowledge of piecewise functions, specifically max functions
  • Basic concepts of real analysis
NEXT STEPS
  • Study the properties of non-decreasing functions in real analysis
  • Explore the application of bounds in integral calculus
  • Learn about piecewise functions and their implications in mathematical proofs
  • Investigate advanced techniques in bounding integrals involving multiple functions
USEFUL FOR

Mathematicians, students of calculus, and researchers in real analysis who are interested in integral bounds and the behavior of functions under specific conditions.

zetafunction
Messages
371
Reaction score
0
given [x] the 'integer function' would be the following bound valid ??

[tex]\int_{0}^{\infty}dt [g(x/t)]f(t) \le \int_{0}^{\infty}dt g(x/t)f(t)[/tex]

for a given functions f(t) and g(u) u=x/t , here g is a non-decreasing positive function for positive arguments (and real) of parameter u=x/t
 
Physics news on Phys.org
Hi,

if f(t) >= 0 for t >=0, then using (g(x/t) + 1) would be a better idea ?
Now if
[tex]f(t) = f^{+}(t) - f^{-}(t)[/tex],
where [tex]f^{+}(t) and f^{-}(t)[/tex] denotes max(f(t),0) and max(-f(t),0) respectively
then one can use g(x/t) + 1 for f+(t) and g(x/t) - 1 for f-(t) to obtain a bound.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K