How can the Lp Norm be used to prove inequalities?

[ashah99]

TL;DR

I would appreciate some help on proofs involving the Lp norm below, please. I looked up Holder's inequality, which might help with this problem, but I'm not sure how to set up the proofs.

 

[anuttarasammyak]

(a_1+a_2+a_3)^\alpha \ge a_1^\alpha+(a_2+a_3)^\alpha \ge a_1^\alpha + a_2^\alpha + a_3^\alpha
a_1+a_2+a_3 \ge [a_1^\alpha + a_2^\alpha + a_3^\alpha]^{\frac{1}{\alpha} }
and so on.

Bonus
f(x,y):=(x+y)^a-x^a-y^a
x,y \ge 0 ,\ \ a \ge 1
\frac{\partial f}{\partial x} \ge 0,\ \ \frac{\partial f}{\partial y} \ge 0, \ \ f(0,0)=0

 

[ashah99]

(a_1+a_2+a_3)^\alpha \ge a_1^\alpha+(a_2+a_3)^\alpha \ge a_1^\alpha + a_2^\alpha + a_3^\alpha
a_1+a_2+a_3 \ge [a_1^\alpha + a_2^\alpha + a_3^\alpha]^{\frac{1}{\alpha} }
and so on.

Bonus
f(x,y):=(x+y)^a-x^a-y^a
x,y \ge 0 ,\ \ a \ge 1
\frac{\partial f}{\partial x} \ge 0,\ \ \frac{\partial f}{\partial y} \ge 0, \ \ f(0,0)=0

How would I prove part (1)? By inspection, the l1 norm will be greater that the lp form when p > 1, since l1 is just the sum of the magnitudes...just not sure how to show work that is proof-based.

Are your hints regarding parts 2 & 3?

 

[anuttarasammyak]

I stopped at 3 but you should continue upto N
(a_1+a_2+...+a_N)^\alpha \ge a_1^\alpha + a_2^\alpha + ... +a_N^\alpha
a_1+a_2+...+a_N \ge (a_1^\alpha + a_2\alpha + ... +a_N^\alpha)^\frac{1}{\alpha}
Let $a_k=|x_k|$

 

[ashah99]

I stopped at 3 but you should continue upto N
(a_1+a_2+...+a_N)^\alpha \ge a_1^\alpha + a_2^\alpha + ... +a_N^\alpha
a_1+a_2+...+a_N \ge (a_1^\alpha + a_2\alpha + ... +a_N^\alpha)^\frac{1}{\alpha}
Let $a_k=|x_k|$

I see that, how does this relate to part 1? I’m slow at understanding it seems.

 

[anuttarasammyak]

(|x_1|+|x_2|+...+|x_N|)^p \ge |x_1|^p +|x_2|^p + ... +|x_N|^p
l_1\ norm=|x_1|+|x_2|+...+|x_N| \ge (|x_1|^p +|x_2|^p + ... +|x_N|^p)^\frac{1}{p}=l_p \ norm

 

[ashah99]

(|x_1|+|x_2|+...+|x_N|)^p \ge |x_1|^p +|x_2|^p + ... +|x_N|^p
l_1\ norm=|x_1|+|x_2|+...+|x_N| \ge (|x_1|^p +|x_2|^p + ... +|x_N|^p)^\frac{1}{p}=l_p \ norm

Okay. this makes sense. Thanks. How would part 2 be proved? I'm thinking I would need to choose some a_i and α and by using the convexity of the function f = x^α. Any thoughts here?

 

[anuttarasammyak]

I'm thinking I would need to choose some a_i and α and by using the convexity of the function f = x^α.

Why don’t you do it ?

Also the hint may be helpful for 2 also. Try substituting $a_k$ to what ?

 

[StoneTemplePython]

How would I prove part (1)? By inspection, the l1 norm will be greater that the lp form when p > 1, since l1 is just the sum of the magnitudes...just not sure how to show work that is proof-based.

Are your hints regarding parts 2 & 3?

since you know the Lp norm is a norm, you should be able to recognize that (1) is implied by triangle inequality, i.e.

$\big \Vert \mathbf x \big \Vert_p=\big \Vert \sum_{k=1}^n x_k \cdot \mathbf e_k\big \Vert_p\leq \sum_{k=1}^n \big \Vert x_k\cdot \mathbf e_k \big \Vert_p = \sum_{k=1}^n \big \Vert x_k\cdot \mathbf e_k \big \Vert_1 = \big \Vert \mathbf x \big \Vert_1$

 

[Office_Shredder]

For (2), if $q<p$, raise both sides to the $q$ power. Then you can pick a clever choice of $y$ to rewrite the inequality as $||y||_{p/q} \leq ||y||_1$.