The scaling argument or "dimensional analysis"

[gugo82]

This easy exercise shows how a simple technique can be used to prove that certain relations (e.g. an inequality) cannot hold for all members of a chosen function space.

This technique is usually called scaling argument or (following the physicists) dimensional analysis: it relies on the comparison of homogeneity properties of the quantities involved in the relations under investigation.

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Some definitions:

The class [tex]$C_c^\infty (\mathbb{R}^N)$[/tex] is made up of all the functions [tex]$u\in C^\infty (\mathbb{R}^N)$[/tex] which are zero outside a compact set; in other words:

[tex]$u\in C_c^\infty (\mathbb{R}^N) \quad \Leftrightarrow \quad u \in C^\infty (\mathbb{R}^N)\ \text{and}\ \exists K\subset \mathbb{R}^N\ \text{compact}:\ \forall x\notin K,\ u(x)=0 $[/tex];

note that [tex]$u=0$[/tex] is the only constant function in [tex]$C_c^\infty (\mathbb{R}^N)$[/tex].

For [tex]$p\in ]0,+\infty[$[/tex] and [tex]$u\in C_c^\infty (\mathbb{R}^N)$[/tex] let:

[tex]$\lVert u\rVert_p := \left\{ \int_{\mathbb{R}^N} |u(x)|^p\ \text{d} x\right\}^\frac{1}{p}$[/tex],

[tex]$\lVert \text{D} u\rVert_p := \left\{ \int_{\mathbb{R}^N} |\text{D} u(x)|^p\ \text{d} x\right\}^\frac{1}{p}$[/tex]*;

the two RHSides are both finite (for the integrals can be thought as Riemann integrals of two continuous functions, both extended over the compact set [tex]$K$[/tex]).
We call [tex]$\lVert u\rVert_p$[/tex] the [tex]$L^p$[/tex] norm of [tex]$u$[/tex] and [tex]$\lVert \text{D} u\rVert_p$[/tex] the [tex]$L^p$[/tex] norm of the gradient [tex]$\text{D}u$[/tex].


__________
* Remember that [tex]$\text{D} u(x) \in \mathbb{R}^N$[/tex], hence [tex]$|\text{D} u(x)|:=\sqrt{\sum_{n=1}^N \left( \frac{\partial u}{\partial x_n}(x)\right)^2}$[/tex].

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Exercise:

Let [tex]$N\in \mathbb{N}$[/tex], [tex]$p \in ]0,+\infty[$[/tex], [tex]$q\in [1,N[$[/tex] and [tex]$q^*:=\tfrac{Nq}{N-q}$[/tex].

Prove that if [tex]$p\neq q^*$[/tex] then there cannot exists any universal constant [tex]$C>0$[/tex] s.t. the inequality (of Sobolev type):

[tex]$\lVert u\rVert_p \leq C\ \lVert \text{D} u \rVert_q$[/tex]

holds for all [tex]$u\in C_c^\infty (\mathbb{R}^N)$[/tex].

Hints: Argue by contraddiction: suppose that the ratio [tex]$\mathcal{R}:=\tfrac{\lVert \text{D} u \rVert_q}{\lVert u\rVert_p}$[/tex] is bounded from below in [tex]$C_c^\infty$[/tex] (here [tex]$u\neq 0$[/tex]).
Fix [tex]$v\in C_c^\infty $[/tex], consider the family of functions defined by [tex]$u_\lambda (x)=v(\tfrac{1}{\lambda}\ x)$[/tex] with [tex]$\lambda >0$[/tex] and evaluate [tex]$\mathcal{R}[u_\lambda]$[/tex] (as a function of [tex]$\lambda$[/tex] and [tex]$\mathcal{R}[v]$[/tex]); derive a contraddiction letting [tex]$\lambda$[/tex] increase/decrease properly.


Btw, the ratio [tex]$\mathcal{R}$[/tex] is sometimes called Rayleigh quotient.

[dissonance]

I have found an easy-to-read and entertaining explanation of this technique in Sanjoy Mahajan's book Street Fighting Mathematics (follow the "Download with Creative Commons License" link). Dimensional analysis makes up for the first chapter.

[Leonardo891]

"gugo82":I write all my post without any WYSIWYG support but the "Anteprima" button.

Wow! And also with the tex tag!

[gugo82]

@Leonardo89: Ok! Correct.

By the way, you're right: actually there is an error in the text!
The correct one is the following:

Let [tex]$N\in \mathbb{N}$[/tex], [tex]$q\in [1,N[$[/tex] and [tex]$p\in ]0,+\infty[$[/tex].

Prove that if [tex]$p\neq q^*:=\tfrac{Nq}{N-q}$[/tex] then there cannot exist any constant [tex]$C>0$[/tex] s.t. the inequality (of Sobolev type):

[tex]$\lVert u\rVert_p \leq C\ \lVert \text{D} u \rVert_q$[/tex]

holds for all [tex]$u\in C_c^\infty (\mathbb{R}^N)$[/tex].

I'm sorry; when I was writing down the exercise I was really distracted by other problems...

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"Leonardo89":Apropos of the problem, Gugo, did you write all of it manually typing the code or did you use a WYSIWYG software?

I write all my post without any WYSIWYG support but the "Anteprima" button.


P.S.: The exponent [tex]$q^*$[/tex] is called the Sobolev conjugate of [tex]$q$[/tex]. By definition, it is the number s.t. [tex]$\tfrac{1}{q} =\tfrac{1}{q^*} +\tfrac{1}{N}$[/tex].

[Leonardo891]

My try...

[tex]$\lVert u_{\lambda} \rVert_p = \left\{ \int_{\mathbb{R}^N} |u_{\lambda} (x)|^p\ \text{d} x\right\}^\frac{1}{p}=\left\{ \int_{\mathbb{R}^N} |v \left( \frac{x}{\lambda} \right) | ^p\ \text{d} x\right\}^\frac{1}{p}$[/tex]

By a change of variables $x= \lambda y $

[tex]$\lVert u_{\lambda} \rVert_p =\left\{ \int_{\mathbb{R}^N} |v \left( y \right) |^p\ \lambda^N \ \text{d} y\right\}^\frac{1}{p}=\lambda^{ \frac{N}{p} } \left\{ \int_{\mathbb{R}^N} |v \left( y \right) |^p\ \text{d} y\right\}^\frac{1}{p}= \lambda^{ \frac{N}{p}} \lVert v \rVert_p$[/tex]

[tex]$\lVert \text{D} u_{\lambda} \rVert_q = \left\{ \int_{\mathbb{R}^N} | \text{D} u_{\lambda} (x)|^q\ \text{d} x\right\}^\frac{1}{q}=\left\{ \int_{\mathbb{R}^N} | \text{D} v \left( \frac{x}{\lambda} \right) | ^q\ \text{d} x\right\}^\frac{1}{q}=$[/tex]

For the theorem of differentiation of composite functions

[tex]$=\left\{ \int_{\mathbb{R}^N} | \left( ( \text{D} v ) \left( \frac{x}{\lambda} \right) \right) \frac{1}{\lambda} | ^q\ \text{d} x\right\}^\frac{1}{q}=\left\{ \int_{\mathbb{R}^N} | \left( ( \text{D} v ) \left( \frac{x}{\lambda} \right) \right) | ^q\ \ \left( \frac{1}{\lambda} \right)^q \ \text{d} x\right\}^\frac{1}{q}$[/tex]

By the same change of variables $x= \lambda y $

[tex]$\lVert \text{D} u_{\lambda} \rVert_q =\left\{ \int_{\mathbb{R}^N} | \left( ( \text{D} v ) \left( y \right) \right) | ^q\ \ \left( \frac{1}{\lambda} \right)^q \ \lambda^N \ \text{d} y\right\}^\frac{1}{q}= \lambda^{ \frac{N}{q}-1} \lVert \text{D} v \rVert_q $[/tex]

Then [tex]$\mathcal{R}[u_{\lambda}]=\tfrac{\lVert \text{D} u_{\lambda} \rVert_q}{\lVert u_{\lambda} \rVert_p}=\tfrac{\lVert \text{D} v \rVert_q }{\lVert v \rVert_p} \lambda^{ \frac{N(p-q)-qp}{qp}} $[/tex] where [tex]$\tfrac{\lVert \text{D} v \rVert_q }{\lVert v \rVert_p} \in \mathbb{R} $[/tex] is fixed.

If $\frac{N(p-q)-qp}{qp}>0$, [tex]$\lim_{\lambda \to 0} \mathcal{R}[u_{\lambda}]=\lim_{\lambda \to 0} \tfrac{\lVert \text{D} v \rVert_q }{\lVert v \rVert_p} \lambda^{ \frac{N(p-q)-qp}{qp}} = 0 $[/tex] therefore [tex]$\mathcal{R}[u_{\lambda}]$[/tex] isn't bounded from below in [tex]$C_c^\infty$[/tex] and this is absurd.

If $\frac{N(p-q)-qp}{qp}<0$, [tex]$\lim_{\lambda \to \infty} \mathcal{R}[u_{\lambda}]=\lim_{\lambda \to \infty} \tfrac{\lVert \text{D} v \rVert_q }{\lVert v \rVert_p} \lambda^{ \frac{N(p-q) -qp}{qp}} = 0 $[/tex] therefore [tex]$\mathcal{R}[u_{\lambda}]$[/tex] isn't bounded from below in [tex]$C_c^\infty$[/tex] and this is absurd.

If $\frac{N(p-q)-qp}{qp}=0$ there's a problem...

Apropos of the problem, Gugo, did you write all of it manually typing the code or did you use a WYSIWYG software?

[gugo82]

I don't know why no one tried to solve this exercise... I think it's easier than any other one i posted before.

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If you're in trouble with partial derivatives, then try to solve at least the onedimensional version:

Exercise (onedimensional version):

Let [tex]$p\neq q \in ]0,+\infty[$[/tex].
Prove that there cannot exists a constant [tex]$C>0$[/tex] s.t. the inequality:

[tex]$\lVert u\rVert_p\leq C\ \lVert u^\prime \rVert_q$[/tex]*

holds true for all [tex]$u\in C_c^\infty (\mathbb{R})$[/tex].

Hints: The same as before.

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Come on...


__________
* Remember that:

[tex]$\lVert u\rVert_p:=\left\{ \int_{-\infty}^{+\infty} |u(x)|^p\ \text{d} x\right\}^\frac{1}{p}$[/tex] and [tex]$\lVert u^\prime \rVert_q:=\left\{ \int_{-\infty}^{+\infty} |u^\prime (x)|^q\ \text{d} x\right\}^\frac{1}{q}$[/tex]

and that the two integrals can be thought as Riemann integrals of continuous functions vanishing outside a suitable compact set.