Answer to Question #153363 in Functional Analysis for Eva

Question #153363

a) What is proposition?

b.) P and q are propositions given by p: f is an odd function and q: βˆ«π‘“(π‘₯)𝑑π‘₯=0π‘˜βˆ’π‘˜

verify whether or not π‘β†”π‘ž


1
Expert's answer
2021-02-24T07:33:42-0500
  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 pβ†’qp \rightarrow q. Suppose that ff is an odd function. Strictly speaking, a proposition q has a meaning only if βˆ«βˆ’kkf(x)dx\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 βˆ«βˆ’kkf(x)dx\int_{-k}^k f (x) dx makes sense, the implication pβ†’qp \rightarrow q is true, as βˆ«βˆ’kkf(x)dx=∫0kf(x)dx+βˆ«βˆ’k0f(x)dx=0\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 qβ†’pq \rightarrow p depends on the class of ff : if it is continuous, then it is true as F(k)=βˆ«βˆ’kkf(x)dx=0,Fβ€²(x)=f(x)+f(βˆ’x)=0F(k) = \int_{-k}^k f(x)dx=0, F'(x) = f(x)+f(-x)=0 and thus f(x)=βˆ’f(βˆ’x)f(x) = -f(-x), it is odd. If we make no hypothesis on ff, then it is obviously false (e.g. f(0)=1,f(x)=0,xβ‰ 0f(0)=1, f(x)=0, x\neq 0) . Therefore for continuous functions this equivalence is true : βˆ€f∈C0(R),βˆ€k∈R,f(βˆ’x)=βˆ’f(x)β†”βˆ«βˆ’kkf=0\forall f \in \mathcal{C}^0 (\mathbb{R}), \forall k\in \mathbb{R}, f(-x)=-f(x) \leftrightarrow \int_{-k}^k f=0.

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