find sup of the function

Question

$$ \sum_{i=1}^n a_ix_i - x_iln(x_i) $$

$a_i$ are given real nos
variables- $x_i$ are all non-negative and $\sum_{i=1}^n = 1$

given is a function from $\mathcal R^n $ to $\mathcal R $ find its sup (it is magically given that $x_iln(x_i)=0$ for $x_i=0 $ )

I am lost, I know concepts of sup and inf...this is a real analysis problem so please avoid using optimization techniques, I want to know the thought process and not just the solution like what made you think in a way that you did.

Answer 1

I assume, you have defined $$ f:\{x\in\mathbb R^n~:~x_i\geq 0\text{ for all }i=1,\ldots,n\}\to \mathbb R $$ defined by $$ f(x)=\sum_{i=1}^na_ix_i-x_i\ln(x_i). $$ The assumption to say $x_i\ln(x_i)\mid_{x_i=0}=0$ is not magical because you can compute that there holds $$ \lim_{t\to 0^+}t\cdot\ln(t)=0. $$ Next, the function looks complicated, but it can be broken down to a simple situation. Let us define $$ g_a:[0,\infty)\to\mathbb R,~g_a(t)=\begin{cases}at-t\cdot\ln(t) & t>0\\0 & t=0\end{cases}. $$ Then, we get $f(x)=\sum_{i=1}^n g_{a_i}(x_i)$.

If you can understand how to compute the supremum and infimum of $g_a$ for an $a\in\mathbb R$, then you can use that to find the supremum and infimum of $f$.

Hints:

If $g_a(t)\to \infty$ then $g_a$ is unbounded from above and $\sup_{t\geq 0} g_a(t)=\infty$. Analogous for $g_a(t)\to-\infty$...

Consider the critical points of $g_a$. They might be useful to find the supremum and infimum of $g_a$. But be aware of the sign of $a$!

$$\sup_{x_1,x_2\geq 0}g_{a_1}(x_1)+g_{a_2}(x_2)=\sup_{x_1\geq 0}g_{a_1}(x_1)+\sup_{x_2\geq 0}g_{a_2}(x_2)$$