Maximizing f(x) with Inequality Constraint: Solving a Functional Inequation

[utkarshakash]

Homework Statement

The function f satisfies $\dfrac{f(x)}{f(y)} \leq 2^{(x-y)^2} x,y \in D$ where D denotes domain set of the function, then f(x) can be

I have a set of options as well but I'm not posting it now. I will post it if required, later.

The Attempt at a Solution

I have dealt with functional equations but this seems more daunting as it is rather an inequation. First, to simplify it, I take natural logarithm of both sides. Then

$log f(x) - logf(y) \leq (x-y)^2 log2$

My next thought is to differentiate the expression wrt x. But I'm not sure whether it would be helpful as I don't want to do it uselessly.

 

[pasmith]

Homework Statement

The function f satisfies $\dfrac{f(x)}{f(y)} \leq 2^{(x-y)^2} x,y \in D$ where D denotes domain set of the function, then f(x) can be

I have a set of options as well but I'm not posting it now.

"Can be", rather than "is", suggests that checking which of those options satisfies this condition is the way forward.

 

[utkarshakash]

"Can be", rather than "is", suggests that checking which of those options satisfies this condition is the way forward.

So how should I check the given options?

 

[pasmith]

So how should I check the given options?

Calculate $f(x)/f(y)$ for each case, and check whether the result is less than or equal to $2^{(x-y)^2}$.

 

[Mark44]

One other thing is that, since (x - y)² ≥ 0 for any real x and y, it must be true that f(x)/f(y) ≥ 2⁰ = 1. You didn't show what the options are, but if there are any that are less than 1, you can eliminate them from further consideration.

Edit: Never mind on the above. I was looking at the wrong direction of the inequality in post #1.

BTW, we don't call it an "inequation" - we call it an inequality.

 

[pasmith]

One other thing is that, since (x - y)² ≥ 0 for any real x and y, it must be true that f(x)/f(y) ≥ 2⁰ = 1.

I'm not sure how that follows from $\dfrac{f(x)}{f(y)} \leq 2^{(x-y)^2}$.

 

[Mark44]

Good point. I must have gotten my inequality sign turned around.

 

[utkarshakash]

Calculate $f(x)/f(y)$ for each case, and check whether the result is less than or equal to $2^{(x-y)^2}$.

For example, one option is

$\int_0^x 2t^3 dt$

This function is x^4/2. You are saying to simply plug x in one and y in another. Doing that gives

$\left( \dfrac{x}{y} \right) ^4$

Now how do I check whether this is less than $2^{(x-y)^2}$ or not? Should I start substituting some random values?

 

[haruspex]

You don't say what D is. Is it specified separately for each f option?
Since the right-hand side of the inequation becomes weak when x and y are far apart, and the inequation is trivially true when x = y, I would concentrate on y and x differing by a small amount.

 

[Ray Vickson]

For example, one option is

$\int_0^x 2t^3 dt$

This function is x^4/2. You are saying to simply plug x in one and y in another. Doing that gives

$\left( \dfrac{x}{y} \right) ^4$

Now how do I check whether this is less than $2^{(x-y)^2}$ or not? Should I start substituting some random values?

You can look at the function
$$F(x,y) = \left(\frac{x}{y}\right)^4 - 2^{(x-y)^2},$$
and try to maximize it, to see if its maximum is ≤ 0.