Reorderings for multivariable infinitums

20 Views Asked by At

I'm working on this optimization problem involving Lagrange Duals and I've run into a problem involving infinitums over multiple variables whose solution contains the equation below. $$g( \lambda ) = {inf}_{x,y>0}(e^{-x} + \lambda \frac{x^2}{y}) = 0 \hspace{20 px} \lambda > 0$$ I'm not sure how they get the right hand side of the equality where the parameter is positive. When I try to think about this conceptually, if I minimize y first, I take it to infinity making the second term x^2 term 0, after which minimizing x gives me x = infinity. Minimizing x first makes my head hurt, but I'm pretty sure it gives me something where the exponential term is non-zero since taking x towards infinity obviously blows up the second term. This suggests I'm thinking about it all wrong since I should just be looking for some point or whatnot that actually makes sense when I plug it in. I know it's the limits that are causing me problems here, but I'm not sure how. I suspect I'm missing something basic (or have some basic misunderstanding of how these sorts of manipulations work), but I'm not sure what it is. Help is appreciated!

1

There are 1 best solutions below

1
On BEST ANSWER

I think it might me misleading thinking about that problem in terms of somewhat "computationally minimizing" the - lets call it function - $f(x, y)=e^{-x} + \lambda x^2 / y$. But what is obvious is, that for $x, y, \lambda >0$ it is always $f(x, y) > 0$, making $0$ the perfect candidate to be the infimum. Now all that is left to do is to find suitable $x$ and $y$ where $f$ acutally approaches $0$. The exponential part already shows, that this can only be achieved by sending $x$ to infinity. Combining this with the second part $y$ has to be send to infitinty as well.

So far for the intuition.

Now proceeding to a more formal solution: Let us consider 2 series $x_n$ and $y_n$ with $x_n \underset{n \to \infty}{\longrightarrow} \infty$ and $y_n \underset{n \to \infty}{\longrightarrow} \infty $ but $x_n^2 / y_n \underset{n \to \infty}{\longrightarrow} 0$. Such series exist with anb easy example beeing $x_n = n$ and $y_n = n^3$. Combined we have two series which fulfill $x_n, y_n >0$ for all $n\in \mathbb N$ and

$$\lim_{n\to \infty} f(x_n, y_n) = 0 \, .$$

That sould show that the infimum is indeed $0$. I hope this helps.

Additionally I would suggest you to use single $ marks for inline math expressions, that would make your post easier to read.