Inequality from a continuity exercise

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I am reading from Courant's book. He gave an example of the continuity of ##f(x)=5x+3## by finding ##\delta=\epsilon/5##. He then said that ##|x-x_0|## does not exceed ##|y-y_0|/5##, but I don't see how he came up with this inequality.
I know that ##|x-x_0|<\epsilon/5##, and that ##|y-y_0|<\epsilon\Leftrightarrow|y-y_0|/5<\epsilon/5##, but I don't think that that suffices to conclude ##|x-x_0|\leq|y-y_0|/5##.
... indeed ##|x-x_0|## is sufficiently small if it does not exceed one-fifth of the value of ##|y-y_0|##.
Any thoughts? Thanks.
 

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  • #2
PeroK
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I am reading from Courant's book. He gave an example of the continuity of ##f(x)=5x+3## by finding ##\delta=\epsilon/5##. He then said that ##|x-x_0|## does not exceed ##|y-y_0|/5##, but I don't see how he came up with this inequality.
I know that ##|x-x_0|<\epsilon/5##, and that ##|y-y_0|<\epsilon\Leftrightarrow|y-y_0|/5<\epsilon/5##, but I don't think that that suffices to conclude ##|x-x_0|\leq|y-y_0|/5##.

Any thoughts? Thanks.
I'm not sure what you are doing here. Can you check what you have written, as it looks confused?

What you want to show is that ##|x - x_0| < \epsilon/5 \ \Rightarrow \ |f(x) - f(x_0)| < \epsilon##. And, given the linearity of ##f## that looks like straightforward?
 
  • #3
Math_QED
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It is unclear from your post what ##y_0## is. I'm guessing you want to show that ##f## is continuous at ##x_0## and ##f(x_0) = y_0##?
 
  • #4
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@PeroK @Math_QED no, I don't want to prove ##f##'s continuity, rather I wanted to verify the author's claim that:
... indeed ##|x-x_0|## is sufficiently small if it does not exceed one-fifth of the value of ##|y-y_0|##.
I can see that ##|x-x_0|=|y-y_0|/5##, and this makes the phrase "is sufficiently small if it does not exceed" a bit confusing. The equality always hold, no matter how small is ##|x-x_0|##, so the only way I could make sense of this is if the author is trying to say that ##|x-x_0|\leq|y-y_0|/5## is true for ##f## to be continuous.
 
  • #5
PeroK
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@PeroK @Math_QED no, I don't want to prove ##f##'s continuity, rather I wanted to verify the author's claim that:

I can see that ##|x-x_0|=|y-y_0|/5##, and this makes the phrase "is sufficiently small if it does not exceed" a bit confusing. The equality always hold, no matter how small is ##|x-x_0|##, so the only way I could make sense of this is if the author is trying to say that ##|x-x_0|\leq|y-y_0|/5## is true for ##f## to be continuous.
Perhaps he hasn't phrased it very well. If you forget what the author says, can you see for yourself how the continuity of ##f## is proved?
 
  • #6
Math_QED
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@PeroK @Math_QED no, I don't want to prove ##f##'s continuity, rather I wanted to verify the author's claim that:

I can see that ##|x-x_0|=|y-y_0|/5##, and this makes the phrase "is sufficiently small if it does not exceed" a bit confusing. The equality always hold, no matter how small is ##|x-x_0|##, so the only way I could make sense of this is if the author is trying to say that ##|x-x_0|\leq|y-y_0|/5## is true for ##f## to be continuous.
$$|x-x_0| <\delta \implies |y-y_0| = |f(x)-f(x_0)| = |5x+3-(5x_0+3)| = 5|x-x_0|< 5\delta$$

This suggest taking ##\delta = \epsilon/5## in an ##\epsilon-\delta## proof.
 
  • #7
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Perhaps he hasn't phrased it very well. If you forget what the author says, can you see for yourself how the continuity of ##f## is proved?
Yes, I do. ##|f(x)-f(x_0)|=5|x-x_0|<\epsilon##, and so we have ##\delta=\epsilon/5##.
$$|x-x_0| <\delta \implies |y-y_0| = |f(x)-f(x_0)| = |5x+3-(5x_0+3)| = 5|x-x_0|< 5\delta$$

This suggest taking ##\delta = \epsilon/5## in an ##\epsilon-\delta## proof.
But this doesn't show that ##|x-x_0|\leq|f(x)-f(x_0)|/5## for ##|x-x_0|<\delta##, which is what I am asking about. I really do understand the continuity proof, trust me!
 
  • #8
PeroK
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I really do understand the continuity proof, trust me!
Why worry if the author got into a verbal tangle? It happens.
 
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  • #9
Math_QED
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But this doesn't show that ##|x-x_0|\leq|f(x)-f(x_0)|/5## for ##|x-x_0|<\delta##
I'm not sure what you are after:
$$\frac{|f(x)-f(x_0)|}{5} = \frac{5|x-x_0|}{5}= |x-x_0|$$

and the inequality you are looking for so hard is always true (it is even equality!) regardless of the fact that ##|x-x_0|## is small.

I agree with @PeroK that the author was probably sloppy and yes it happens. I'm currently reading an analysis book full of mistakes.
 
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Thank you both. @Math_QED and @PeroK
Why worry if the author got into a verbal tangle? It happens.
Well, I am still learning, and when I come upon stuff like this, I cannot help but feel frustrated for not understanding. And I don't really trust my knowledge enough at this point to judge by myself that such and such is a typo/mistake or not.
 
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