This entry devotes the following fundamental question: if is Hölder continuous, then how about for some constant ? Throughout this entry, we work on which is not necessarily bounded.
Firstly, we have an elementary result
Proposition. If and are -Hölder continuous and bounded, so is .
Proof. The proof is simple, we just observe that
for any positive integer number and any -Hölder continuous and bounded function , function is also -Hölder continuous and bounded.
Let us assume , is -Hölder continuous and bounded, is a constant. Let . Since , we may assume is also bounded away from zero, that means there exist two constants such that
We now study the -Hölder continuity of . Observe that function
is sub-additive in the sense that
Indeed, by dividing both sides by we arrive at
the above inequality comes from the Bernoulli inequality. Therefore if there are some then
holds for any . We now arrive at
We now make use the above estimate to show that is -Hölder continuous. Indeed, we have the following
Consequently, we get
we claim that is -Hölder continuous.
Let us now turn to the case . For the sake of simplicity, let us denote , i.e. the function under investigating is . Obviously
Thus is -Hölder continuous since
for any . Consequently,
for any positive integer number and any -Hölder continuous and bounded function , function is -Hölder continuous.
Similarly, we can prove
is -Hölder continuous. It is now easy to prove that is also -Hölder continuous. So far we have proved the following
Theorem. If is -Hölder continuous, positive and bounded (i.e. ). We assume in addition is bounded away from zero (i.e. ). Then is also -Hölder continuous for any constant .
The boundedness of plays an important role in our argument. In practice, if strictly positive function is locally -Hölder continuous on , we are the able to apply the theorem. We will discuss something related to this application later.