So we know that a function $f(x)$ is continuous at a point $x_0$ if for each $\epsilon >0$ we can find a $\delta$ s.t. $\mid x_0-x\mid \lt \delta \implies \mid f(x) - f(x_0) \mid \lt \epsilon $. I was wondering if there is any definition which uses multiplication with $x$ meaning that for $x$ very close to $1$ we can make $\mid f(x_0.x) - f(x_0) \mid \lt \epsilon $. I was thinking maybe uniform continuity could imply something like this. Thank you in advance!
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$|x-h| < \delta$? – Gustavo Louis G. Montańo Feb 09 '15 at 01:14
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1One problem with your proposed definition is that every function defined at $x=0$ would be continuous at $0$. – paw88789 Feb 09 '15 at 01:14
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I am not saying that argument work both ways I just try to find if it works for continuous functions. – SpawnKilleR Feb 11 '15 at 03:49
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If $f:>x\mapsto f(x)$ is continuous at $x_0$ then $p\mapsto f(px_0)$ is continuous at $1$. – Christian Blatter Mar 18 '16 at 09:23
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Except for $x=0$ that will work equivalently. You simply have if you want $x_0-x<\delta$ you could equally well say that $x = x_0\xi$ where $\xi$ is sufficiently close to $1$ since $x-x_0 = (\xi-1)x_0$.
Suppose that first you have the first definition and $\epsilon>0$ then we have a $\delta$ so that $|x_0-x|<\delta\Rightarrow |f(x)-f(x_0)|<\epsilon$. Now suppose that $|\xi-1|<\delta/|x_0|$, then we have that $|x_0-\xi x_0|=|x_0||1-\xi| < |x_0|\delta/|x_0| = \delta$, and therefore we have $|f(x_0)-f(\xi x_0)|<\epsilon$.
Similar reasoning gets the other way around that if we have a $\eta$ such that $|\xi-1|<\eta\Rightarrow|f(\xi x_0)-f(x_0)|<\epsilon$, that $|\xi x_0 - x_0| = |x_0||\xi-1| < \eta |x_0|$. So we have a $\delta = \eta|x_0|$ such that $|x-x_0|<\delta \Rightarrow |f(x)-f(x_0)|<\epsilon$
The question is how useful the alternative definition is. First of all it's quite useless for $x_0=0$, but otherwise?
skyking
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