How to show that for a continuous function on $\mathbb{R}$ that satisfies $-f(-x)=f(x)$ for all $x \neq0$ then $f(0)=0$

Question

How to show that for a continuous function on $\mathbb{R}$ that satisfies $-f(-x)=f(x)$ for all $x \neq0$ then $f(0)=0$

For $-f(-x)=f(x)$ to be true, it appears we must have an odd function without a constant at the end. You can tell this from examination (it's obvious for a linear function, but should also hold true for cubic functions and upward). So it seems obvious to me that the proposal is true, but I don't know of a theorem to prove my observation, or to move from my observation to an actual proof. Any help?

Answer 1

$$f(-x)=-f(x)， \forall x \ne 0$$

$$\lim_{x \to 0}f(-x)=\lim_{x \to 0}-f(x)$$

By continuity, $$f(\lim_{x \to 0}-x)=-f(\lim_{x \to 0}x)$$

$$f(0)=-f(0)$$

$$2f(0)=0$$

$$f(0)=0$$

Answer 2

With an elementary "$\varepsilon$-$\delta$" proof.

By continuity, we have $f(0) = \lim_{x\to 0} f(x)$. We will use the definition of limit.

Fix any $\varepsilon > 0$; and let $\delta_\varepsilon > 0$ be such that $$ \forall x \neq 0 \text{ s.t. } |x|\leq \delta_\varepsilon, \, |f(x)-f(0)| \leq \varepsilon. \tag{1} $$ In particular, fix any $x\in (0,\delta_\varepsilon]$. We have, applying the above to both $x$ and $-x$, $$ |f(x)-f(0)| \leq \varepsilon \;\text{ and }\; |f(x)+f(0)| = |-f(x)-f(0)| = |f(-x)-f(0)| \leq \varepsilon \tag{2} $$ so that (using the triangle inequality) $$ 2|f(0)| = |f(0)-f(x)+f(x)+f(0)| \leq |f(0)-f(x)|+|f(x)+f(0)| \leq 2\varepsilon \tag{3} $$ i.e., $|f(0)| \leq \varepsilon$. Since the above works for every $\varepsilon > 0$, this implies $|f(0)|=0$, i.e., $f(0)=0$.

Answer 3

Consider the sequence $$\{1/n\}$$

Due to continuity of $f(x)$ we have $$ \lim _{n\to \infty} f(1/n) =f(0)$$

Similarly we have $$ \lim _{n\to \infty} f(-1/n) =f(0)$$

Since we have $$f(1/n)=-f(-1/n)$$ We have

$$ f(0)= -f(0)$$

Thus $$f(0)=0$$

Answer 4

Suppose $f(0)>0$. The case $f(0)<0$ is analogous.

Let $\varepsilon=\dfrac{f(0)}{2}>0$.

How $f$ is continuous at $0$ (in particular), there exists $\delta>0$ such that $|x-0|<\delta \Rightarrow |f(x)-f(0)|<\varepsilon$.

So, $f(x)\in \left( f(0)-\varepsilon, f(0)+\varepsilon \right) = \left( \dfrac{f(0)}{2}, \dfrac{3\cdot f(0)}{2} \right)$

$\Rightarrow f(x)>\dfrac{f(0)}{2}>0$, $\forall x \in (-\delta, \delta)$.

But, if $f(x)>0$ for some $x\in(-\delta,\delta)$, then $f(-x)<0$ with $-x\in(-\delta,\delta)$. Contradiction!

Therefore $f(0)=0$

Answer 5

Define $\, g(x) := f(x) + f(-x). \,$ As a sum of continuous functions, $\, g(x) \,$ is continuous and $\, g(x) = 0 \,$ for all $\, x \neq 0. \,$ By continuity, $\, g(0) = f(0) + f(-0) = 2f(0) = 0. \,$ Now $\, f(0) = 0.$