Epsilon Delta Proof Piecewise function

[Dethrone]

https://answers.yahoo.com/question/index?qid=20130915100124AAK4JAQ

I do not understand how they got:
"1 = |(1 plus d/2 - L) - (d/2 - L)| <= |1 plus d/2 - L| plus |d/2 - L| < 1/4 plus 1/4 = 1/2, "

Shouldn't it be $|(1+ \frac{\delta}{2} -L) + (\frac{\delta}{2} -L)|$, not $|(1+ \frac{\delta}{2} -L) - (\frac{\delta}{2} -L)|$

 

[Euge]

https://answers.yahoo.com/question/index?qid=20130915100124AAK4JAQ

I do not understand how they got:
"1 = |(1 plus d/2 - L) - (d/2 - L)| <= |1 plus d/2 - L| plus |d/2 - L| < 1/4 plus 1/4 = 1/2, "

Shouldn't it be $|(1+ \frac{\delta}{2} -L) + (\frac{\delta}{2} -L)|$, not $|(1+ \frac{\delta}{2} -L) - (\frac{\delta}{2} -L)|$

No, $|(1 + \frac{\delta}{2} - L) + (\frac{\delta}{2} - L)| = |1 + \delta - 2L|$, not $1$. Adding and subtracting $\frac{d}{2} - L$ gives

\(\displaystyle 1 = 1 + (\frac{\delta}{2} - L) - (\frac{\delta}{2} - L) = (1 + \frac{\delta}{2} - L) - (\frac{\delta}{2} - L)\),

and thus

\(\displaystyle 1 = |(1 + \frac{\delta}{2} - L) - (\frac{\delta}{2} - L)|\).

 

[Dethrone]

I agree, but how was he able to continue with the triangle inequality?

I thought it was $|x+y| \le |x| + |y|$, so if they rewrote it as $|x-y|$, how is $|x-y|\le |x|+|y|$?

 

[mathbalarka]

They're trying to derive a contradiction. I presume you have followed the steps they took in order to deduce $\left | \delta/2 + 1 - L \right | < 4$ from the hypothesis of having a limit?

What they did next was sheer trickery. Note that $(1 + \delta/2 - L) - (\delta/2 - L) = 1$, thus,

$$1 = \left | \left(1 + \frac{\delta}{2}-L \right) - \left ( \frac{\delta}{2} - L \right) \right | \leq \left | 1 + \frac{\delta}{2} - L \right | +\left | \frac{\delta}{2} - L \right |$$

Which follows from the triangle ineq. But then we have deduced that

$$\left | 1 + \frac{\delta}{2} - L \right | +\left | \frac{\delta}{2} - L \right | \leq \frac14 + \left | 1 + \frac{\delta}{2} - L \right | < \frac14 + \frac14 = \frac12$$

Combining the two above, $1 < 1/2$, a contradiction.

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I agree, but how was he able to continue with the triangle inequality?

I thought it was $|x+y| \le |x| + |y|$, so if they rewrote it as $|x-y|$, how is $|x-y|\le |x|+|y|$?

$$|x-y| = |x + (-y)| \leq |x| + |-y| = |x| + |y|$$

 

[Dethrone]

Oh okay! It's just that my professor has another formula for $|x-y|$ that is different, but it matters not.

Suppose we want to show that the Heaviside function does not exist at $t=0$.

Heaviside function is the following:
$$H(t)=\begin{cases}1, & t\ge 0 \\[3pt] 0, & t<0 \\ \end{cases}$$

Following the same method, assume that $\epsilon = 1/2$ is imposed, such that we want to find $0<|t|<\delta$ such that $|H(t) -L|<1/2$. If we pick $t =\delta /2$ which satisfies, we have $|1-L|<1/2$, similarly picking $t=-\delta / 2$, we have $|-L|<1/2$.

By the triangle inequality, we find that:
$1=|1-L-(-L)|\le |1-L|+|L| < 1/2 + 1/2 = 1$

Contradiction because 1 is not less than 1. Is that correct? Do I have to say anything about $\delta$ = min{${\text{something}}$}?

 

[mathbalarka]

That looks good to me. (Yes)

 

[Euge]

Oh okay! It's just that my professor has another formula for $|x-y|$ that is different, but it matters not.

You mean the reverse triangle inequality, $||x| - |y|| \le |x - y|$?

 

[Dethrone]

Exactly, but are those external absolute values necesary? My prof left it out, he said for any real numbers $x$ and $y$:
$|x-y|\ge |x|-|y|$

 

[Euge]

What I have implies your inequality, but keep in mind that $|x| - |y|$ may be negative.