Prove Inequality: ||x|^α - |y|^α| ≤ |x-y|^α

[IniquiTrance]

Homework Statement

Prove the following inequality holds:

$||x|^\alpha - |y|^\alpha | \leq |x-y|^\alpha \qquad (\forall x,y\in \mathbb{R}, \alpha \in (0,1])$

Homework Equations

The Attempt at a Solution

I tried squaring both sides, getting:$x^{2 \alpha} - 2 (|x||y|)^\alpha + y^{2 \alpha} \leq (x^2 - 2xy + y^2)^\alpha$

Any help is much appreciated.

 

[scurty]

Okay, your first mistake is that you are assuming what you are trying to prove.

I'll start you off and see what you can do from there. This is a two step proof.

Start with $|x|^{\alpha}$ as one case and $|y|^{\alpha}$ as the other. Your goal is to arrive at a conclusion similar to the one you are trying to prove and then combine your answers to get the final answer.

Here's the hint: x = x - y + y and y = y - x + x. See what you can come up with!

 

[IniquiTrance]

Alrighty, thanks for the hint! Please tell me if the following is correct:

By the triangle inequality:

$d(y,x+y) \leq d(y, x) + d(x, x+y)$
$(d(y,x+y))^\alpha \leq (d(y,x) + d(x, x+y) )^\alpha \leq (d(y,x))^\alpha + (d(x, x+y))^\alpha$
$\therefore (d(y,x+y))^\alpha - (d(x, x+y))^\alpha \leq (d(y,x))^\alpha$

Similarly,

$(d(x, x+y))^\alpha - (d(y,x+y))^\alpha \leq (d(x,y))^\alpha$

Proving that $|(d(x, x+y))^\alpha - (d(y,x+y))^\alpha| \leq (d(x,y))^\alpha$.

 

[scurty]

You switched notation up on me, but it looks good. Don't forget to include the part where $d(x,y)=d(y,x)$.

 

[IniquiTrance]

You switched notation up on me, but it looks good. Don't forget to include the part where $d(x,y)=d(y,x)$.

Haha, sorry about that. Thanks for the help!

 

[IniquiTrance]

Hmm, after looking things over again, I realized I'm making an unjustified assumption.

Rewriting my proof in our original notation, slightly differently:


$|x-0| \leq |x-y| + |y-0|$
$|x|^\alpha \leq \left(|x-y| + |y|\right)^\alpha \leq |x-y|^\alpha + |y|^\alpha$
$|x|^\alpha - |y|^\alpha \leq |x-y|^\alpha$

So on the second line, last expression, I have assumed sub-additivity. Namely,

$\left(|x-y| + |y|\right)^\alpha \leq |x-y|^\alpha + |y|^\alpha$

I know how to prove this is true for $\alpha = 1/2, 1$, but how can I prove it is true for all $\alpha\in (0,1]$?

Thanks!

 

[scurty]

Consider $f(x) = x^{\alpha}, \quad \alpha \in (0,1], \quad x \in \mathbb{R}^{+}$. (this is because in the problem we are considering positive values only, no need for negatives)

What can you tell me about the derivative of $f\prime (x)$? Something about the derivative that relates $x_{1} < x_{2}$ to $x_{1}^{\alpha} < x_{2}^{\alpha}$.

 

[IniquiTrance]

Hmm, since $f'(x)=\alpha x^{\alpha - 1}$, and $\alpha - 1 < 0$ for $\alpha \in (0,1)$, we have that f ' (x) is decreasing in x. So since x+y>x, and x+y>y by assumption, we have f(x+y)<f(x) and f(x+y)<f(y). But why does this imply f(x+y)<= f(x) + f(y)? Also, why is there no loss of generality by assuming both x,y>0?

Edit: Hmm, what I said above doesn't appear to be right. f '(b) < f '(a) does not necessarily imply f (b)<f(a). I think I've confused myself even further.

 

[scurty]

f'(x) is actually increasing, just because the exponent is negative doesn't mean the number is negative.

What I am getting at is that since $x \geq 0$, $f(x) = x^{\alpha}, \quad \alpha \in (0,1]$ is an increasing function. So for $a<b, \quad f(a) = a^{\alpha} \leq b^{\alpha} = f(b)$. I'm sorry if I confused you. If the function was decreasing, $a<b$ would imply $\quad f(a) = a^{\alpha} \geq b^{\alpha} = f(b)$ which is not what we are working towards. Does that makes sense?

 

[IniquiTrance]

Yes, that was rather careless on my part, not your fault at all. I'm not sure what the next step would be, though.

 

[IniquiTrance]

Ah ok, would it be something along the lines of this, then?

We wish to show that $(|x-y|+|y|)^\alpha \leq |x-y|^\alpha +|y|^\alpha \quad \alpha\in (0,1]$. This is equivalent to showing $(a+b)^\alpha \leq a^\alpha + b^\alpha \quad a,b >0, \alpha\in (0,1].$

We want to show now that the derivative of the LHS is smaller than the derivative of the RHS on (0,1].

Differentiating both sides WRT alpha yields:

$\alpha (a+b)^{\alpha -1} \ ? \ \alpha (a^{\alpha - 1} + b^{\alpha - 1})$
$(a+b)^{\alpha -1} \ ? \ a^{\alpha - 1} + b^{\alpha - 1}$

Where ? represents the relation to be determined. Now $\alpha - 1 <0$. Therefore,

$1 \ ? \ (\frac{a}{a+b})^{\alpha - 1} + (\frac{b}{a+b})^{\alpha - 1}$
$1 \ ? \ (\frac{a+b}{a})^{1-\alpha} + (\frac{a+b}{b})^{1 - \alpha}$
$1 \ ? \ (1+ \frac{b}{a})^{1-\alpha} + (1 + \frac{a}{b})^{1 - \alpha}$

Therefore $? = <$ for $\alpha\in (0,1)$.

Does this seem right?

 

[scurty]

I don't think a derivative is going to help us here. We used it before to show if $a \leq b$, then $f(a) \leq f(b)$. Derivatives involve functions, in that case we considered $f(x) = x^{\alpha}$ where x was a positive real number and $\alpha \in (0,1]$.

I'm a little bit stumped right now so I did a bit of searching and found this http://en.wikipedia.org/wiki/Subadditivity . This would help in our case but I doubt this is what is required how to solve this. I'm currently stumped, hopefully someone else can chime in, sorry! I'll think about it some more in the mean time!

 

[IniquiTrance]

Edit: Woops. Derivatives would actually be $\ln[a+b](a+b)^\alpha$ and $\ln[a]a^\alpha, \ln<b>b^\alpha</b>$, which as you say would not be of much help.

 

[IniquiTrance]

Ok so a friend of mine proved it as follows:

Assume for the sake of contradiction that $x\neq y$ and:

$||x|^\alpha - |y|^\alpha |>|x-y|^\alpha$
$\Leftrightarrow ||x|^\alpha - |y|^\alpha |^{1/ \alpha} >|x-y|$

Taking limits as alpha -> 0 from above:

$0=\lim_{\alpha \rightarrow 0^+} ||x|^\alpha - |y|^\alpha |^{1/ \alpha} >|x-y|>0 \qquad \blacksquare$