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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.

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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 $|f(x)-f(-x_0)| <\epsilon$.

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  • $\begingroup$ Perfect answer. $\endgroup$ – Sam Streeter Oct 26 '18 at 9:45
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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 $g(x)=-x$.

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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

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

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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 $-x_0$.

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