I have a function
$$f(x)=(K-x)^2 + \left( \frac{T}{x} \right)^2$$
where $K$ and $T$ are positive constants and $x>0$.
The function $f$ (hopefully) has an infimum, in terms of $K$ and $T$.
I want to put an upper bound for this infimum, only as a function of $T$. And this function should go to zero, as $T$ goes to zero.
in other words, I want to show that
$$\lim_{T \to 0^+} \sup_{K \in \mathbb R^+} \inf_{x \in \mathbb R^+} f(x) =0$$
Actually, I don't know whether it is true or not. Can you please help me for this problem?
I appreciate any help. Thanks.
Let $x=\sqrt{T}+K$. Then $(K-x)^2=T$ and $\left(\frac{T}{x}\right)^2\lt T$. So the minimum is $\le 2T$ for any $K$. The supremum over all positive $K$ of the minima is indeed exactly $2T$.