Answer to Question #153363 in Functional Analysis for Eva

Question

a) What is proposition?

b.) P and q are propositions given by p: f is an odd function and q: ∫𝑓(𝑥)𝑑𝑥=0𝑘−𝑘

verify whether or not 𝑝↔𝑞

Answer 1

1. A proposition, from a mathematical logic viewpoint, is a logical expression without free variables. From a viewpoint of "general" mathematics it is a statement meant to be proven.
2. First let's look at $p \rightarrow q$. Suppose that $f$ is an odd function. Strictly speaking, a proposition q has a meaning only if $\int_{-k}^k f (x) dx$ exists (i.e. f integrable), as we can have an odd function that is not Riemann-integrable/Lebesgue-integrable etc. Assuming that the expression $\int_{-k}^k f (x) dx$ makes sense, the implication $p \rightarrow q$ is true, as $\int_{-k}^k f(x)dx = \int_0^k f(x)dx+\int_{-k}^0 f(x)dx =0$ by changing the variables in the second integral. The other direction $q \rightarrow p$ depends on the class of $f$ : if it is continuous, then it is true as $F(k) = \int_{-k}^k f(x)dx=0, F'(x) = f(x)+f(-x)=0$ and thus $f(x) = -f(-x)$, it is odd. If we make no hypothesis on $f$, then it is obviously false (e.g. $f(0)=1, f(x)=0, x\neq 0$) . Therefore for continuous functions this equivalence is true : $\forall f \in \mathcal{C}^0 (\mathbb{R}), \forall k\in \mathbb{R}, f(-x)=-f(x) \leftrightarrow \int_{-k}^k f=0$.