You are correct. For a simple counterexample, take ##f(x) = 0## for all ##x##. Then ##f \in L^2##, but if ##c## is any nonzero constant, then ##f + c \not\in L^2##, for the reason you indicated.lmedin02 said:Homework Statement
Suppose [itex]f\in L^2(\mathbb{R}^2)[/itex]. Is [itex]f+c\in L^2(\mathbb{R}^2)[/itex] where c is a constant?
Homework Equations
[itex]f\in L^2(\mathbb{R}^2)[/itex] if [itex]||f||_2<∞[/itex].
The Attempt at a Solution
I think the answer is no because [itex]∫_{\mathbb{R}^2}{c^2}dx=∞[/itex]. However, I am still unsure. Any guidance is appreciated.
[itex]L^2(\mathbb{R}^2)[/itex] is a function space that contains all square integrable functions defined on [itex]\mathbb{R}^2[/itex]. In other words, a function [itex]f(x,y)[/itex] belongs to [itex]L^2(\mathbb{R}^2)[/itex] if the integral of [itex]|f(x,y)|^2[/itex] over [itex]\mathbb{R}^2[/itex] is finite.
To show that a function [itex]f(x,y)[/itex] is in [itex]L^2(\mathbb{R}^2)[/itex], you need to prove that the integral of [itex]|f(x,y)|^2[/itex] over [itex]\mathbb{R}^2[/itex] is finite. This can be done by using techniques such as change of variables, integration by parts, or applying known inequalities.
Functions in [itex]L^2(\mathbb{R}^2)[/itex] have many useful properties, such as being square integrable and having a well-defined Fourier transform. These properties make them useful in many areas of mathematics and physics, including signal processing, quantum mechanics, and probability theory.
Yes, it is possible for a function to be in [itex]L^2(\mathbb{R}^2)[/itex] but not in [itex]L^1(\mathbb{R}^2)[/itex]. This means that the function is square integrable but not necessarily integrable. In other words, the integral of [itex]|f(x,y)|^2[/itex] may be finite, but the integral of [itex]|f(x,y)|[/itex] may not be.
[itex]L^2(\mathbb{R}^2)[/itex] is an example of a Hilbert space, which is a complete vector space with an inner product defined on it. In [itex]L^2(\mathbb{R}^2)[/itex], the inner product is given by [itex]\langle f,g \rangle=\int_{\mathbb{R}^2}f(x,y)\overline{g(x,y)}\,dx\,dy[/itex]. This makes [itex]L^2(\mathbb{R}^2)[/itex] a powerful tool for studying functions and solving problems in a rigorous mathematical framework.