The discussion confirms that if a function f belongs to L²(ℝ²), then adding a constant c results in f + c not belonging to L²(ℝ²) if c is non-zero. This is established through the counterexample where f(x) = 0 for all x, which is in L²(ℝ²), but f + c does not satisfy the condition ||f + c||₂ < ∞ due to the integral of c² over ℝ² being infinite. Thus, the conclusion is definitive: f + c ∉ L²(ℝ²) for any non-zero constant c.
PREREQUISITESMathematics students, particularly those studying functional analysis, as well as educators and researchers interested in the properties of L² spaces and Lebesgue integration.
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.