# Prove an inequality

1. Mar 12, 2012

### IniquiTrance

1. The problem statement, all variables and given/known data
Prove the following inequality holds:

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

2. Relevant equations

3. 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.

Last edited: Mar 12, 2012
2. Mar 12, 2012

### 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!

3. Mar 12, 2012

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

4. Mar 12, 2012

### scurty

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

5. Mar 12, 2012

### IniquiTrance

Haha, sorry about that. Thanks for the help!

6. Mar 13, 2012

### 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!

7. Mar 13, 2012

### 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 $x_{1}^{\alpha} < x_{2}^{\alpha}$.

8. Mar 13, 2012

### 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.

9. Mar 13, 2012

### 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?

10. Mar 13, 2012

### 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.

11. Mar 13, 2012

### 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?

12. Mar 13, 2012

### 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!

13. Mar 13, 2012

### IniquiTrance

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

Last edited: Mar 13, 2012
14. Mar 13, 2012

### 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:

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