Let be a continous and bounded function such that for all

Then is constant.

Advertisements

Skip to content
# Constant Function

##
5 thoughts on “Constant Function”

### Leave a Reply

Let be a continous and bounded function such that for all

Then is constant.

Advertisements

%d bloggers like this:

Because is continuous and integrals of continuous functions are differentiable, it is evident that is smooth, and for every . In fact, for every , so that if is a bound for , is a bound for . This proves that must be analytic, and that in fact that . But then defines an holomorphic function on all of and [TODO].

LikeLiked by 2 people

Just an aside on whether the boundedness is needed or not: as noted above, any as in the statement satisfies , which is a linear delay differential equation (DDE) with constant coefficients and discrete delays with associated characteristic equation (see for instance section 1.3 of this thesis). A bound for may probably be obtained from this.

LikeLiked by 1 person

Obviously there are non-bounded solutions, like , but they do not satisfy the original integral equation…

LikeLike

Here is an argument relying on the Fourier transform which shows that the only bounded solutions for the associated DDE are the constant ones.

LikeLike

Recall that the Laplace transform of a continuous bounded function is defined as , and this defines an operator where is the space of continuous bounded functions over the positive reals. Furthermore, this operator is an isomorphism of Banach spaces.

Integrating by parts, it is easy to see that if is a continuously differentiable function such that is bounded, then . It is also easy to see that if then .

Now, let be a function that satisfies the hypothesis. WLOG we can assume that . Since the hypothesis implies that for all , by applying the Laplace transform we obtain for all , and thus . Since it is an isomorphism for bounded functions, we get for . By noticing that satisfies the hypothesis we obtain for all . And we are done.

LikeLiked by 1 person