Consider arbitrary $0 < \lambda \leqslant n$, $1 \leqslant p < \infty$ and $\alpha,\beta > 0$ such that $\alpha < \frac{n-\lambda}{p} < \beta.$ My goal is to compute the supremum of the function
$$ f(r) = r^{-\lambda}\left[ \frac{c(n)}{-\alpha p + n} + k(n)\left( \frac{r^{-\beta p + n}}{-\beta p + n} - \frac{1}{-\beta p + n}\right)\right]$$
where $c(n)$ and $k(n)$ are unknown positive constants, over $ r \geqslant 1$.
My attempt. Since $ \alpha < \frac{n-\lambda}{p}$, we know that $-\alpha p + n > \lambda > 0,$ which implies that the term $c(n)/(-\alpha p + n)$ is positive. Therefore, $r^{-\lambda}\frac{c(n)}{-\alpha p + n}$ attains its supremum exactly when $r = 1$. On the other hand, the other terms are quite hard to analyze:
- First, the inequality $\frac{n-\lambda}{p} < \beta$ doesn't give me any conclusion on the signal of $-\beta p + n$, which complicates the calculations;
- On the other hand, we know that $-\beta p + n -\lambda < 0,$ which implies that the supremum of $r^{-\beta p + n - \lambda}$ is attained when $r=1$;
- If we assume that $-\beta p + n > 0,$ then the supremum of $k(n)\frac{r^{-\beta p+n-\lambda}}{-\beta p+n}$ is also attained when $r=1$, just from what I wrote on the last point;
- Now, if we keep assuming that $-\beta p + n>0$, then the sumpremum of $-k(n)\frac{r^{-\lambda}}{-\beta p + n}$ is attained when $r \to \infty$.
Clearly, from the points I stated above it is impossible to reach any conclusion. So, I am looking for a new way of approaching this problem. Ideally, I am looking for a solution that doesn't assume anything about $-\beta p + n$, but I strongly believe this is not possible, so a solution that assumes that $-\beta p + n > 0$ will be good enough. Furthermore, if you have any other suggestion of assumptions (perhaps changing the initial inequalities to make the exercise easier) I am open to listen to them.
Thanks for any help in advance.
We can write
$$ f(r) = \frac{c}{(n-\alpha p)r^{\lambda}} + \frac{k}{(n-\beta p)r^{\beta p-n+\lambda}}-\frac{ k}{(n-\beta p)r^{\lambda}} $$ We have $f(1)\gt 0$ and $\displaystyle\lim_{r\to\infty}f(r)=0$.
With the help of WolframAlpha, we have $$f'(r)=\frac{ g(r)}{(n - \alpha p) (n - \beta p)r^{\beta p-n+2\lambda+1}}$$ and $$f''(r)=\frac{h(r) }{(n - \alpha p) (n - \beta p)r^{\beta p-n+2\lambda +2}}$$ where $$g(r):=Ar^{\lambda}-Br^{\beta p-n+\lambda}$$
$$h(r):=Cr^{\lambda} - Dr^{\beta p-n+\lambda}$$
$$A:=k(\alpha \beta p^2 + \alpha \lambda p - \alpha n p - \beta n p - \lambda n + n^2 )$$
$$B:=\lambda(\alpha k p - \beta c p - k n + c n)$$
$$C:=k( 2 \beta \lambda n p - 2 \beta n^2 p + \beta n p + \lambda^2 n - 2 \lambda n^2 + \lambda n + n^3 - n^2 -\alpha \beta^2 p^3 - 2 \alpha \beta \lambda p^2 - \alpha \beta p^2 + 2 \alpha \beta n p^2 - \alpha \lambda^2 p - \alpha \lambda p + 2 \alpha \lambda n p - \alpha n^2 p + \alpha n p + \beta^2 n p^2)$$
$$D:=\lambda(-c \lambda n + \beta c \lambda p - c n - \alpha k p +\beta c p + k \lambda n - \alpha k \lambda p + k n )$$
Case 1 : If $n-\beta p\gt 0$, then $\lambda\gt \beta p-n+\lambda$, so $$g(r)=0\iff Ar^{\lambda}-Br^{\beta p-n+\lambda}=0\iff Ar^{n-\beta p}-B=0\iff r^{n-\beta p}=\frac{B}{A}$$ Let $r_1:=(\frac{B}{A})^{1/(n-\beta p)}$ when $\frac{B}{A}\geqslant 0$.
We have $f(r)\gt 0$.
If $\frac{B}{A}\lt 0$, then $f(1)$ is the maximum.
If $\frac{B}{A}\geqslant 0$ and $r_1\lt 1$, then $f(1)$ is the maximum.
If $\frac{B}{A}\geqslant 0$ and $r_1\geqslant 1$, then $f(r_1)$ is the maximum.
Case 2 : If $n-\beta p\lt 0$, then $\lambda\lt \beta p-n+\lambda$, so $$g(r)=0\iff Ar^{\lambda}-Br^{\beta p-n+\lambda}=0\iff A-Br^{\beta p-n}=0\iff r^{\beta p-n}=\frac{A}{B}$$ Here, note that $(\frac{A}{B})^{1/(\beta p-n)}=r_1$.
If $\frac{A}{B}\lt 0$, then $f(1)$ is the maximum.
If $\frac{A}{B}\geqslant 0$ and $r_1\lt 1$, then $f(1)$ is the maximum.
If $\frac{A}{B}\geqslant 0,r_1\geqslant 1$ and $f''(r_1)\lt 0$, then $f(r_1)$ is the maximum.
If $\frac{A}{B}\geqslant 0,r_1\geqslant 1$ and $f''(r_1)\gt 0$, then $f(1)$ is the maximum.
Added 1 :
I realized that we can have a simple conclusion where we don't have to find $f''(r)$.
We have $f(r)\gt 0$ since $$f(r) = \underbrace{\frac{cr^{-\lambda}}{n-\alpha p}}_{\text{positive}} + \underbrace{kr^{-\lambda}}_{\text{positive}}\times \underbrace{\frac{r^{n-\beta p}-1}{n-\beta p}}_{\text{non-negative}}\gt 0$$ We also have $A\lt 0$ since $$A=\underbrace{k(\beta p+\lambda-n)}_{\text{positive}}\times\underbrace{(\alpha p-n)}_{\text{negative}}\lt 0$$
Conclusion :
$$\color{red}{\text{max}f(r)=\begin{cases}f(1)&\text{if $(n-\beta p)(A-B)\lt 0$} \\f(r_1)&\text{otherwise}\end{cases}}$$
Added 2 :
If $n-\beta p\gt 0$, then $$f'(r)=\underbrace{\frac{Ar^{-\lambda-1}}{(n - \alpha p) (n - \beta p)}}_{\text{negative}}\bigg(r^{n-\beta p}-\frac BA\bigg)$$
$(1)$ If $\frac{B}{A}\lt 0$, then since $f'(r)\lt 0$, we see that $f(1)$ is the maximum.
$(2)$ If $\frac{B}{A}\geqslant 0$ and $r_1\lt 1$, then since $f'(r)\lt 0$, we see that $f(1)$ is the maximum.
$(3)$ If $\frac{B}{A}\geqslant 0$ and $r_1\geqslant 1$, then $f(r_1)$ is the maximum.
If $n-\beta p\lt 0$, then $B=\lambda k(\alpha p-n)+\lambda c(n-\beta p)\lt 0$, so $\frac AB\gt 0$. We have $$f'(r)=\underbrace{\frac{Br^{-\beta p+n-\lambda-1}}{(n - \alpha p) ( \beta p-n)}}_{\text{negative}}\bigg(r^{\beta p-n}-\frac AB\bigg)$$
$(4)$ If $r_1\lt 1$, since $f'(r)\lt 0$, we see that $f(1)$ is the maximum.
$(5)$ If $r_1\geqslant 1$, then $f(r_1)$ is the maximum.
We can say the followings :
There are only five cases $(1), (2),\cdots, (5)$.
Each of $(1)(2)(4)$ has $(n-\beta p)(A-B)\lt 0$ and $\text{max}f(r)=f(1)$.
Each of $(3)(5)$ has $(n-\beta p)(A-B)\geqslant 0$ and $\text{max}f(r)=f(r_1)$.
There is no case which has $(n-\beta p)(A-B)\lt 0$ and $\text{max}f(r)=f(r_1)$.
There is no case which has $(n-\beta p)(A-B)\geqslant 0$ and $\text{max}f(r)=f(1)$.
So, we can say the followings :
If $(n-\beta p)(A-B)\lt 0$, then $\text{max}f(r)=f(1)$.
If $(n-\beta p)(A-B)\geqslant 0$, then $\text{max}f(r)=f(r_1)$.
Therefore, we get the above conclusion in red.
Added 3 :
We don't need the assumption. We can prove that.
(1) We have $$f'(r)=\underbrace{\frac{Ar^{-\lambda-1}}{(n - \alpha p) (n - \beta p)}}_{\text{negative}}\times\underbrace{\bigg(r^{n-\beta p}-\frac BA\bigg)}_{\text{positive}}\lt 0$$
$(2)$ We have $0=r_1^{n-\beta p}-\frac BA\lt 1-\frac BA\leqslant r^{n-\beta p}-\frac BA$, so $$f'(r)=\underbrace{\frac{Ar^{-\lambda-1}}{(n - \alpha p) (n - \beta p)}}_{\text{negative}}\times\underbrace{\bigg(r^{n-\beta p}-\frac BA\bigg)}_{\text{positive}}\lt 0$$
$(3)$ For $1\leqslant r\lt r_1$, since $r^{n-\beta p}-\frac BA\lt r_1^{n-\beta p}-\frac BA=0$, we have $$f'(r)=\underbrace{\frac{Ar^{-\lambda-1}}{(n - \alpha p) (n - \beta p)}}_{\text{negative}}\times\underbrace{\bigg(r^{n-\beta p}-\frac BA\bigg)}_{\text{negative}}\gt 0$$ For $r_1\lt r$, since $0=r_1^{n-\beta p-n}-\frac BA\lt r^{n-\beta p}-\frac BA$, we have $$f'(r)=\underbrace{\frac{Ar^{-\lambda-1}}{(n - \alpha p) (n - \beta p)}}_{\text{negative}}\times\underbrace{\bigg(r^{n-\beta p}-\frac BA\bigg)}_{\text{positive}}\lt 0$$
$(4)$ We have $0=r_1^{\beta p-n}-\frac AB\lt 1-\frac AB\leqslant r^{\beta p-n}-\frac AB$, so $$f'(r)=\underbrace{\frac{Br^{-\beta p+n-\lambda-1}}{(n - \alpha p) ( \beta p-n)}}_{\text{negative}}\times\underbrace{\bigg(r^{\beta p-n}-\frac AB\bigg)}_{\text{positive}}\lt 0$$
$(5)$ For $1\leqslant r\lt r_1$, since $r^{\beta p-n}-\frac AB\lt r_1^{\beta n-n}-\frac AB=0$, we have $$f'(r)=\underbrace{\frac{Br^{-\beta p+n-\lambda-1}}{(n - \alpha p) ( \beta p-n)}}_{\text{negative}}\times\underbrace{\bigg(r^{\beta p-n}-\frac AB\bigg)}_{\text{negative}}\gt 0$$ For $r_1\lt r$, since $0=r_1^{\beta p-n}-\frac AB\lt r^{\beta p-n}-\frac AB$, we have $$f'(r)=\underbrace{\frac{Br^{-\beta p+n-\lambda-1}}{(n - \alpha p) ( \beta p-n)}}_{\text{negative}}\times\underbrace{\bigg(r^{\beta p-n}-\frac AB\bigg)}_{\text{positive}}\lt 0$$