# If the odd function $f:\mathbb R\to\mathbb R$ letting $x>0$ is continuous at $x$, prove the function is continuous at $-x$.

If the odd function $$f :\mathbb R \to\mathbb R$$ letting $$x > 0$$ is continuous at $$x$$, prove the function is continuous at $$-x$$.

I've been trying to do an epsilon delta proof where I let $$f$$ be continuous at some $$x_0>0$$ so that $$\forall\epsilon>0, \exists\delta>0$$ s.t. $$\lvert f(x)-f(x_0)\rvert<\epsilon$$ when $$\lvert x-x_0\rvert<\delta$$ and trying to show that for $$-x_0$$, which I let be $$x_1$$ s.t. $$x_1=-x_0$$,$$\forall\epsilon_1>0, \exists\delta_1>0$$ s.t. $$\lvert f(x)-f(x_1)\rvert<\epsilon_1$$ when $$\lvert x-x_1\rvert<\delta_1$$, but I'm getting stuck beyond this point.

At an abstract level, you could use

If $$g$$ is continuos at $$a$$ and $$f$$ is continuous at $$g(a)$$, then $$f\circ g$$ is continuous at $$a$$

and apply it to $$g(x)=-x$$.

No need for a different $$\epsilon_1$$. Take $$\epsilon_1=\epsilon$$ and take $$\delta_1=\delta$$. If $$|x-(-x_0)| <\delta$$ then $$|(-x) -x_0| <\delta$$ so $$|f(-x) -f(x_0)| <\epsilon$$. Now use the fact that $$f$$ is an odd function to get $$|f(x)-f(-x_0)| <\epsilon$$.

• Perfect answer. Oct 26 '18 at 9:45

Given $$f$$, odd function, continuous a:

$$\epsilon >0$$, there is a $$\delta$$ s.t.

$$|x-(a)| \lt \delta$$

implies

$$|f(x)-f(a)| \lt \epsilon.$$

Rewrite:

$$|(-1)(x -a)| =|(-x) -(-a)| \lt \delta$$ implies

$$|-f(-x)+f(-a)| =$$

$$|f(-x)-f(-a)| \lt \epsilon.$$

Set $$y:=- x$$.

Then

$$|y-(-a)| \lt \delta$$ implies

$$|f(y)-f(-a)| \lt \epsilon.$$

With sequences: let $$f$$ be continuous at $$x_0$$.

If $$(x_n)$$ is a sequence with limit $$-x_0$$, then $$-x_n \to x_0$$, hence $$f(-x_n) \to f(x_0)$$. From $$f(-x_n)=-f(x_n)$$ we get

$$-f(x_n) \to f(x_0)$$.

Therefore $$f(x_n) \to -f(x_0)=f(-x_0)$$. This shows that $$f$$ is continuous at $$-x_0$$.