Regarding Wirtinger's inequality

In summary, "Regarding Wirtinger's inequality" discusses the mathematical concepts and applications of Wirtinger's inequality, which provides bounds on the integral of a function based on its derivatives. The inequality is significant in various fields such as functional analysis and partial differential equations, as it establishes a relationship between the function's average value and its oscillation. The text explores the conditions under which the inequality holds, its generalizations, and implications for the study of Sobolev spaces and other mathematical constructs.
  • #1
psie
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I have a doubt about about an assumption in Wirtinger's inequality.
Consider the following problem:
Let ##f## be a continuous real-valued function on ##0<x<\pi## such that ##f(0)=f(\pi)=0## and ##f' \in L^2([0,\pi])##.
a) Prove that $$\int_0^{\pi}(f(x))^2 dx \leq \int_0^{\pi} (f'(x))^2 dx.$$
b) For what functions does equality hold?
What I struggle with in this exercise is why ##f'## is in ##L^2([0,\pi])##. Does this make sense? The way I prove this inequality is that I extend ##f## to an odd function on ##[-\pi,\pi]##. I find its Fourier series, namely ##f(x)\sim\sum_{n=1}^\infty b_n\sin nx##. Then, since ##f'## will be even, its Fourier coefficients are $$a_n=\frac{2}{\pi}\int_0^\pi f'(x)\cos(nx)dx=\frac{2n}{\pi }\int_0^\pi f(x)\sin(nx)=nb_n,$$ since ##f(0)=f(\pi)=0## by partial integration. Then I use Parseval's formula $$\int_0^\pi f^2= \frac{\pi}{2}\sum_{n=1}^\infty b_n^2 \quad \text{and} \quad \int_0^\pi (f')^2= \frac{\pi}{2}\sum_{n=1}^\infty n^2b_n^2.$$ And the inequality follows from this. Equality holds for ##a_n=0## for ##n\neq 1##, i.e. when ##f(x)=C\sin x##.

In order for this solution to work, we need ##f'## to be integrable, right? Does ##f'\in L^2([0,\pi])## guarantee this?
 
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  • #2
psie said:
In order for this solution to work, we need ##f'## to be integrable, right? Does ##f'\in L^2([0,\pi])## guarantee this?
We need square integrability of ##f'## to even phrase the statement and ##f'\in L^2## is just that.
 
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  • #3
fresh_42 said:
We need square integrability of ##f'## to even phrase the statement and ##f'\in L^2## is just that.
I oversaw something there :smile: In my solution, however, I also need ##a_n=\frac2{\pi}\int_0^\pi f’(x)\cos(nt)dt## to be well-defined. Does ##f’\in L^2## ensure this?
 
  • #4
psie said:
I oversaw something there :smile: In my solution, however, I also need ##a_n=\frac2{\pi}\int_0^\pi f’(x)\cos(nt)dt## to be well-defined. Does ##f’\in L^2## ensure this?
Cauchy-Schwarz
 

FAQ: Regarding Wirtinger's inequality

What is Wirtinger's inequality?

Wirtinger's inequality is a fundamental result in mathematical analysis that provides a bound on the integral of a function in terms of its gradient. Specifically, it states that if a function is square-integrable and has a zero mean over a certain interval, then there exists a constant such that the integral of the square of the function is bounded by the integral of the square of its derivative.

In what contexts is Wirtinger's inequality used?

Wirtinger's inequality is widely used in various fields of mathematics, including functional analysis, partial differential equations, and the theory of Sobolev spaces. It is particularly useful in establishing the existence of solutions to differential equations and in studying the properties of functions in various function spaces.

What are the conditions for Wirtinger's inequality to hold?

For Wirtinger's inequality to hold, the function in question must be sufficiently smooth (typically at least continuously differentiable) and must have a zero mean over the interval of interest. Additionally, the interval must be bounded, and the function must be defined on that interval.

How does Wirtinger's inequality relate to Sobolev spaces?

Wirtinger's inequality is closely related to the theory of Sobolev spaces, as it provides an essential tool for establishing embeddings of these spaces. It shows that the norm of a function in a Sobolev space can be controlled by the norm of its derivative, which is crucial for proving compactness and continuity properties in functional analysis.

Can Wirtinger's inequality be generalized to higher dimensions?

Yes, Wirtinger's inequality can be generalized to higher dimensions. In multiple dimensions, the inequality can be applied to functions defined on bounded domains, and it can involve the gradient in multiple directions. The general form still maintains the essence of relating the integral of the function to the integral of its gradient, but the specifics may vary depending on the geometry of the domain and the boundary conditions imposed.

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