Proving a function f has a maxima given its two asymptotic ends

Question

I have a function $f(x)$ is a continuous differentialfunction in $x>0$ such that $f(x) = C$ for $x \ll x_0$ and $f(x) = \alpha\cdot x^{-\frac{1}{d}}$ for $x \gg x_0$ where $x_0 >0$ is a fixed number, $\alpha >0$ is a constant and $d\in \mathbb{Z}$ and $d >0$. From this information can we prove that $f(x)$ would have a maxima for some x. If not, what other information about $f(x)$ should I know (minimal requirement) for $f(x)$ to have a maxima which is greater than $C$.

Answer 1

Yes, $f$ has a maximum. Here's a sketch of the proof. We have $f(x)=\alpha x^{-1/d}$ for $x \gg x_0$. Since $f'(x)=-\frac{\alpha}{d}x^{-(1/d+1)}<0$ with $f(x)\to 0$ as $x\to\infty$, it follows that for some large $M>0$, we have $$ \max_{x\in[M,\infty)}f(x)=f(M). $$ Then obviously $$ \max_{x\in(0,\epsilon]}f(x)=C, $$ where $\epsilon \ll x_0$. Finally, on $[\epsilon,M]$ the function $f$ is continuous so it achieves a maximum value since $[\epsilon,M]$ is compact. Say $\alpha=\max_{x\in[\epsilon,M]} f(x)$. Then we can say $$ \max_{x\in(0,\infty)}f(x)=\max\{C,\alpha,f(M)\}. $$