Show $|\exp(-x/2) - \exp(-y/2)| \leq |x-y|/4$ for $x,y\geq 0$.

Question

I am trying to show this inequality: $$ \left|e^{-x/2} - e^{-y/2}\right| \leq \frac{|x-y|}{4} $$ for $x,y\geq 0$.

I've gotten stuck and could use some kind assistance.

Many thanks in advanced!

Answer 1

This is incorrect as written. E.g., for $x = 0, y = 1$, $$LHS = 1 - e^{-1/2} \approx 0.393 \color{red}{>} \frac{1}{4} = RHS$$

However it is true that $$\left|e^{-x/2} - e^{-y/2}\right| \leq \frac{|x-y|}{2} \quad\text{ for all } x, y \geq 0$$ To see this, write $f(x) = e^{-x/2}$. Then by the Mean Value Theorem, for all $0 \leq x < y$,

$$\frac{e^{-x/2} - e^{-y/2}}{x-y} = f'(c) \quad\text{ for some } c \in (x,y)$$ The absolute value of the derivative $\left|f'\right|$ is bounded above on the non-negative reals by $1/2$. Hence $$\left|\frac{e^{-x/2} - e^{-y/2}}{x-y}\right| \leq \frac 1 2$$

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One other alternative: if you require the bound to be

$$\left|e^{-x/2} - e^{-y/2}\right| \leq \frac{|x-y|}{4}$$ then this will the case for $x, y \geq a$ such that $\left|f'(a)\right| = \frac 14$, i.e., $x, y \geq 2 \ln 2$.

Answer 2

Hint: In terms of the distance between $x$ and $y$, it is not hard to show that the maximum difference between $e^{-x/2}$ and $e^{-y/2}$ occurs when either $x$ or $y$ is zero, assuming $x,y \geq 0$. Also, keep in mind what the other answer said about the incorrect constant.

Answer 3

Just here to add some intuition.

As @SimonS said, this problem is best managed with the MVT. To have seen this right away, note that we've got an inequality involving the distance between two constants on one side (i.e., $x$ and $y$) and the distance between two manipulations of (mappings of) those same constants on the other side--$e^{-\frac{1}{2}x}$ and $e^{-\frac{1}{2}y}$.

So you should see the inequality $$|e^{-\frac{1}{2}x}-e^{-\frac{1}{2}y}|\leq\frac{1}{2}|x-y|$$ and think:

Does the mapping $e^{-\frac{1}{2}\cdot}$ bring points closer together than they started?

(i.e., the distance on the LHS is smaller than the distance on the RHS, which means the numbers $e^{-\frac{1}{2}x}$ and $e^{-\frac{1}{2}y}$ are closer together than are $x$ and $y$)

Whether a function brings points closer together or not can be directly related to (the absolute value of!) the function's derivative--consider $f_1(x)=x$, $f_2(x)=\frac{1}{2}x$, and $f_3(x)=2x$:

Hopefully it's plain as day that $f_1$ keeps points at the same distance--after all, $f_1$ doesn't change the points. It should be almost as clear that $f_2$ brings points closer together (e.g. $0$ and $1$ are 1 apart, but $f_2(0)$ and $f_2(1)=\frac{1}{2}$ are only $\frac{1}{2}$ apart) and $f_3$ spreads points apart. The situation is the same when the slope is negative--just put a mirror on the $x$-axis.

So if a function has a derivative like $f_2$ it brings points closer together (keeping in mind that all statements about derivatives are, strictly speaking, "local", i.e., the points we start at matter).

Which does our function behave like? The derivative of $e^{-\frac{1}{2}x}$ is $-\frac{1}{2}e^{-\frac{1}{2}x}$, which is negative, and gets closer to 0 as $x$ gets larger. This means that the larger $x$ gets, the closer it comes to behaving like $f_2$.

So, since we're only considering positive $x,y$, and the larger $x$ and $y$ are the closer our mapping gets to making the images of $x$ and $y$ overlap, we need only focus on the behavior at 0 of the mapping.

There, the derivative is $-\frac{1}{2}$, the absolute value of which should look familiar from the question statement. From here, we let the MVT take over.