Proving Holder Continuity for Composite Functions

[rsq_a]

I'd like to show that functions like $$x^a$$ with $$a > 0$$ satisfy the Holder condition on an interval like [0, 1]. That is to say that for any x and y in that interval, then for example,

$$|x^{\frac{1}{2}} - y^{\frac{1}{2}}| \leq C|x-y|^k$$

for some constants C and k.

What is the trick to proving these sorts of things? For Lipschitz continuity, I remember the trick of using mean value theorems with triangle inequalities. And this this?

I'd appreciate some help. It's been a while since I've done analysis.

 

[snipez90]

Well for the specific example you chose with a = 1/2, it's fairly simple. The trick is to realize that
$$|x^{\frac{1}{2}} - y^{\frac{1}{2}}| \leq |x^{\frac{1}{2}} + y^{\frac{1}{2}}|.$$
One way to prove this is to use the fact that |x| = sqrt(x^2) to reduce the inequality to something that is evidently true while ensuring that all of your steps are reversible. Upon multiplying both sides of the inequality by
$$|x^{\frac{1}{2}} - y^{\frac{1}{2}}|$$
and taking square roots, we get the desired Holder condition with C = 1, k = 1/2.

It should be possible to solve the general problem with more work, but I haven't given it too much thought.

 

[rsq_a]

Well for the specific example you chose with a = 1/2, it's fairly simple. The trick is to realize that
$$|x^{\frac{1}{2}} - y^{\frac{1}{2}}| \leq |x^{\frac{1}{2}} + y^{\frac{1}{2}}|.$$
One way to prove this is to use the fact that |x| = sqrt(x^2) to reduce the inequality to something that is evidently true while ensuring that all of your steps are reversible. Upon multiplying both sides of the inequality by
$$|x^{\frac{1}{2}} - y^{\frac{1}{2}}|$$
and taking square roots, we get the desired Holder condition with C = 1, k = 1/2.

It should be possible to solve the general problem with more work, but I haven't given it too much thought.

Thank you. This helped me.

For the general problem, it seems that the trick (after perusing this), is to first establish the property that if $f$ and $g$ are Holder continuous and bounded, then so is their product. Then, instead of looking at $f(x) = x^\gamma$, you can look at x to the power of floors and ceilings of $\gamma$, which allows you to expand brackets and so on.

There are a couple things I don't understand about the proof on that post. The author tries to establish that if $u(x)$ is Holder continuous, then so is $u^\gamma$ for $\gamma>0$ initially. However, he states that since $\gamma \in \mathbb{R}$, then it is sufficient to assume $u(x)$ is bounded away from zero. I don't understand this. Clearly $u^{1/2}$ on [0, 1] is totally different from $u^{1/2}$ on [1, 2]!

Another thing I don't understand is the line right above the line, "we claim that $u^\gamma$ is $\alpha$-Hölder continuous."

It seems that he is using the product

$$u^{\lfloor\gamma\rfloor}u^{\lceil\gamma\rceil}$$,

but its relation to $u^\gamma$ is unclear to me.

 

[bookworm_vn]

There are a couple things I don't understand about the proof on that post. The author tries to establish that if $u(x)$ is Holder continuous, then so is $u^\gamma$ for $\gamma>0$ initially. However, he states that since $\gamma \in \mathbb{R}$, then it is sufficient to assume $u(x)$ is bounded away from zero. I don't understand this. Clearly $u^{1/2}$ on [0, 1] is totally different from $u^{1/2}$ on [1, 2]!

Another thing I don't understand is the line right above the line, "we claim that $u^\gamma$ is $\alpha$-Hölder continuous."

It seems that he is using the product

$$u^{\lfloor\gamma\rfloor}u^{\lceil\gamma\rceil}$$,

but its relation to $u^\gamma$ is unclear to me.

According the theorem in that post, the result is limited to the case when the function $u$ is bounded and bounded away from zero.

 

[bookworm_vn]

Concerning to "we claim that $$u^\gamma$$ is $$\alpha$$-Hölder continuous", this comes from the definition and the fact that the LHS of the inequality on the line right above is finite.

The key point here is to deal with $$f \circ g$$, like

$$u(x)^\alpha = f(u(x))$$ with $$f(t) =t^\alpha$$.