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Can anyone double check my following epsilon delta proof.

I want to prove that the following function is continuous with an Epsilon Delta Argument. $$ f: x \in \mathbb R \mapsto (4x^2+3x+17) \in \mathbb R $$

So I started with $$\left\lvert f(x)-f(y) \right\rvert =$$ $$= \left\lvert (4x^2+3x+17) - (4y^2+3y+17) \right\rvert$$ $$= \left\lvert (4x^2+3x)-(4y^2+3y) \right\rvert$$ $$= 3\left\lvert (4x^2+x)-(4y^2+y) \right\rvert$$ $$= 3\left\lvert (\frac {4}{3}x^2+x)-(\frac {4}{3}4y^2+y) \right\rvert $$ $$= 3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert$$ $$= 3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert$$ (Triangel inequality)

$$\le 3 \left(\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 \right\rvert + \left\lvert x-y \right\rvert \right)$$

Note that $\delta \le 1$

$$\le 3 \left(\frac {4}{3}\left\lvert x^2 - y^2 \right\rvert + \delta \right)$$

$$= 3 \left(\frac {4}{3}\left\lvert (x - y) \right\rvert\left\lvert (x +y) \right\rvert + \delta \right)$$

$\delta \le 1$ $$\le 3 \left(\frac {4}{3}\left\lvert x + y \right\rvert \delta + \delta \right)$$ Adding zero in form of y - y $$= 3 \left(\frac {4}{3}\left\lvert x + y - y + y \right\rvert \delta + \delta \right)$$ $$= 3 \left(\frac {4}{3}\left\lvert x - y + 2y \right\rvert \delta + \delta \right)$$ Triangel inequality $$\le 3 \left(\frac {4}{3}(\left\lvert x - y \right\rvert +\left\lvert 2y \right\rvert) \delta + \delta \right)$$ Note that $\delta \le 1$

$$\le 3 \left(\frac {4}{3}(\delta +\left\lvert 2y \right\rvert) \delta + \delta \right)$$ $$= 4\delta^2+4\left\lvert 2y \right\rvert \delta + 3\delta $$ $$= \delta(4\delta+4\left\lvert 2y \right\rvert + 3) $$ Note that $\delta \le 1$ $$\le \delta(4 +4\left\lvert 2y \right\rvert + 3) $$ $$\le \delta(\left\lvert 8y \right\rvert + 7) = \epsilon$$ Therefore $$\delta = \frac {\epsilon}{(\left\lvert 8y \right\rvert + 7)} \;\; with \; \delta \le1$$

Thanks in advance for the help, I really appreciate it. :)

  • There are so many errors here.. errors or typo – Gabrielek Jan 14 '21 at 17:43
  • As I was typing this comment you corrected your error of $(M_x + N_x) - (M_y + N_y) \ne (M_x + M_y)-(N_x-N_y)$ to $(M_x+N_x)-(M_y+N_y) = (M_x - M_y) -(N_x-N_y)$. But now you have an error $|a^2 - b^2| \ne |a-b||a-b|$. $|a^2 - b^2| = |a+b||a-b|$. – fleablood Jan 14 '21 at 17:53
  • @fleablood Thank you! Those are all typos, I don't know why I have so many errors in my latex code, but on the proof on my paper it is correct facepalm I am so sorry. –  Jan 14 '21 at 17:55
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    Your final line is that $\delta$ is expressed in terms of the variable $y$. That is an absolute and unresolvable no-no. – fleablood Jan 14 '21 at 17:57
  • @fleablood Why is that a problem? (I am a student so.....) –  Jan 14 '21 at 18:01
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    @fleablood his $y$ is (something like) $x+\delta$, so maybe with some modifications between the last two lines it's fixable? – Benjamin Wang Jan 14 '21 at 18:02
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    "Why is that a problem?" Because to choice $y$ so that $|x-y| < \delta$ you have to know what $\delta$ is first. If $\delta = $something to do with $y$ you have to know $y$ first which is circular; you have to know what $y$ is before you can choose it. @BenjaminWang "his y is (something like) x+δ, so maybe with some modifications between the last two lines it's fixable? " Maybe. Probably. Conventionally we don't do $\delta$ dependent on $x$ but.... off hand that will prove it is continuous at that $x$ but as $x$ is arbitrary that could be acceptable... I think. – fleablood Jan 14 '21 at 18:20
  • @fleablood But look for example at this epsilon delta proof for $x^3$: https://math.stackexchange.com/questions/2378786/proof-that-fx-x3-is-continuous Here delta is also dependend on y (or c in the other post) –  Jan 14 '21 at 18:30
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    $|2y| < 2|x| + 2\delta$ so $\delta(8|x| + 4\delta + 7)\le (8|x|+ 4)\delta +7$ ought to do it. I thing. But I'm a bit fussy about saying "Therefore $\delta = ....$" ... that's not really the logic of the proof. Better to say "therefore of we choose a $\delta$ so that $0 < \delta \le \min(..., 1)$ that will be a valid choice and our result will follow" or something like that. – fleablood Jan 14 '21 at 18:31
  • @fleablood I agree with you on that part. But per Definition I only need to proof that there exists a $\delta$. Isn't it okay then? –  Jan 14 '21 at 18:33
  • @fleablood So basically Therefore if we choose delta like _____ then follows abs(f(x)-f(y))< $\epsilon$. That would do it in your opinion, right @fleablood? –  Jan 14 '21 at 18:36
  • That's dependent on $x=c$. Not on $y=x_0$ which is the point that they chose. To prove that a function is continuous at a particular point $x = c$. The $x=c$ is fixed. You have to show that if you pick a variable unfixed $y$ somewhere so that $|y - x| <\delta$ that .... whatever..... Because the *$y$* is what you pick you can't have $\delta$ based on it. But the *$x$* is fixed and steady. You can have $\delta$ based on $x=c$. That one particular $x=c$. For another value of $x=k$ you'd have a different $\delta$ based on a different value $k$. – fleablood Jan 14 '21 at 18:38
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    " Isn't it okay then?" Yes, and that delta will work. And any delta smaller would work. And that is not the largest possible choice of delta. But it is an acceptable delta. To my mind saying "therefore $\delta =$" sounds like you are saying delta must be that value. It'd be more accurate to say "If we choose $\delta$ to be this value that will be sufficient". ... but it's a minor point. – fleablood Jan 14 '21 at 18:45
  • Thank you so much for your help (and sorry again for the typos at the beginning)! I am a statistics student, hence I do not have pure math lectures, so my more "formal proofs" often lack some minor details. If you would write an answer I would check it, but you do not have to. :) –  Jan 14 '21 at 18:51
  • This post might be helpful. – Ivo Terek Jan 14 '21 at 19:39
  • That made things more clear. Thank you @IvoTerek –  Jan 14 '21 at 20:02
  • Why do you want to factor out $3$ at the beginning? It makes things so much more complicated. –  Jan 14 '21 at 21:05
  • @mrsamy In retrospect it really was not that smart. But the idea was to get to the point that I |x-y| by using the triangle inequality –  Jan 14 '21 at 21:10
  • @mrsamy I should have extracted |x-y| and then used the fact that |x|-|y| <= |x-y| < delta and there for. |x| < 1 + |y| (delta). Then I can just solve it more easily but the delta is the same –  Jan 14 '21 at 21:15
  • @Bruno: then you could factor out $4$ instead of $3$ to avoid using fractions :-) –  Jan 14 '21 at 21:16
  • Yeah that is a good point, but it didn't bother me I don't know why haha. –  Jan 14 '21 at 21:17
  • @mrsamy Thank you again so much. I would not know what I would do without you ;-). –  Jan 14 '21 at 21:18
  • @Bruno: you are welcome! –  Jan 14 '21 at 21:19

2 Answers2

1

Your solution looks good. The following steps could be simplified a bit to make it more readable.

\begin{align} \left\lvert f(x)-f(y) \right\rvert &= 3\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 + x-y \right\rvert \quad (\textrm{why bother factoring out $3$?})\\ & \le 3 \left(\left\lvert \frac {4}{3}x^2 - \frac {4}{3}y^2 \right\rvert + \left\lvert x-y \right\rvert \right)\\ & \le 3 \left(\frac {4}{3}\left\lvert x^2 - y^2 \right\rvert + \delta \right) \quad (\color{red}{\textrm{for $|x-y|<\delta$} })\\ &= 3 \left(\frac {4}{3}\left\lvert (x - y) \right\rvert\left\lvert (x +y) \right\rvert + \delta \right)\\ &= 3 \left(\frac {4}{3}\left\lvert x - y + 2y \right\rvert \delta + \delta \right)\\ &\le 3 \left(\frac {4}{3}(\left\lvert x - y \right\rvert +\left\lvert 2y \right\rvert) \delta + \delta \right) \quad (\textrm{triangle inequality})\\ &\le 3 \left(\frac {4}{3}(\delta +\left\lvert 2y \right\rvert) \delta + \delta \right)\\ &=4(\delta +\left\lvert 2y \right\rvert) \delta + 3\delta \\ &\le 4(1+\left\lvert 2y \right\rvert) \delta + 3\delta \quad (\delta\le 1)\\ &\le (7+|8y|)\delta \end{align}

Set $$\delta =\min(\frac {\epsilon}{\left\lvert 8y \right\rvert + 7} ,1).$$

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To prove that $f(x)=4 x^2 + 3 x + 17$ is continuous $\forall a\in\mathbb{R}$

$$\underset{x\to a}{\text{lim}}\left(4 x^2+3 x+17\right)=f(a)$$ Equivalently $$\underset{h\to 0}{\text{lim}}\left(4 (a+h)^2+3 (a+h)+17\right)=4 a^2 + 3 a + 17$$ $\forall \varepsilon>0$ we must find a $\delta>0$ such that if $|h|<\delta$ then $$|4 (a+h)^2+3 (a+h)+17-(4 a^2 + 3 a + 17)|<\varepsilon$$ that is $$|8 a h+4 h^2+3 h|<\varepsilon$$

$$0<\delta<\left|\frac{1}{8} \sqrt{64 a^2+48 a+16 e+9}+\frac38-a\right|$$ if $|h|<\delta$ then $|f(a+h)-f(a)|<\varepsilon$

Raffaele
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  • Thanks for your post. But could you please verify if or if not my answer is correct. I know that there are different ways like yours but it doe not really help me. –  Jan 14 '21 at 20:59
  • Express $\delta$ as $ \frac {\varepsilon}{7}$. So you prove that $f(x)$ is uniformly continuos on any $[a,b]$ – Raffaele Jan 14 '21 at 21:17
  • Why can I just leave out the $ (\left \vert 8y \right \vert )^{-1}$? –  Jan 14 '21 at 21:21