Odd Function of Zero is Zero

Theorem

Let $f: \R \to \R$ be an odd function.

Let $f$ be defined at the point $x = 0$.

Then:

$\map f 0 = 0$

Proof

By definition of odd function:

$\map f {-x} = -\map f x$

and so:

\(\ds \map f {-0}\)

\(=\)

\(\ds \map f 0\)

\(\ds \)

\(=\)

\(\ds -\map f 0\)

The only real number $a$ for which $a = -a$ is $0$.

Hence the result.

$\blacksquare$

Sources

• 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Chapter One: $\S 4$. Even and Odd Functions