Integrating by an other variable

[evinda]

Hello! (Wave)

I am looking at a proof where we have:

$$\int_{-\infty}^{+\infty} |u|^{\frac{n}{n-1}} dx_1 \leq \left( \int_{-\infty}^{+\infty} |Du| dy_1 \right)^{\frac{1}{n-1}} \left( \prod_{i=2}^n \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} |Du| dx_1 dy_i\right)^{\frac{1}{n-1}}$$

Integrating with respect to $x_2$ , I get the following:

$ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} |u|^{\frac{n}{n-1}} dx_1 dx_2 \leq \left( \int_{-\infty}^{+\infty} |Du| dy_1\right)^{\frac{1}{n-1}} \left( \int_{-\infty}^{+\infty} |Du| dy_2\right)^{\frac{1}{n-1}} \int_{-\infty}^{+\infty} \left( \prod_{i=3}^n \int_{-\infty}^{+\infty} \left( \int_{-\infty}^{+\infty} |Du| dy_i\right) dx_1\right)^{\frac{1}{n-1}} dx_3 $.But according to my notes, we get this:View attachment 5618But it doesn't hold that $ \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} |Du| dx_1 dy_2 = \left( \int_{-\infty}^{+\infty}|Du| dy_1 \right)\left( \int_{-\infty}^{+\infty}|Du| dy_2 \right)$.

Does it ?

Why does the product begin from $1$ and not from $3$? Do we just bound the product I got by the product starting from $1$?

 

[Jhoneine]

No, it does not hold that $\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} |Du| dx_1 dy_2 = \left( \int_{-\infty}^{+\infty}|Du| dy_1 \right)\left( \int_{-\infty}^{+\infty}|Du| dy_2 \right)$. The product begins from $1$ because the integral is taken over the entire domain, so all variables must be included in the product. Therefore, the product should start from $i=1$ and end at $i=n$.

 

[math771]

Hello! (Wave)

It looks like your proof may be missing some steps. Can you provide more context or explain how you got to this point? It's hard to say whether or not the product should start at 1 or 3 without understanding the reasoning behind it. And as for the equality, it's possible that it may not hold in general, but it could hold for certain cases depending on the properties of the function and the bounds of integration. Can you provide more information?