# Is my solution correct in terms of proof? I can't prove directly that $f(x)=4$.

Given $$f:\mathbb{R}\rightarrow \mathbb{R}^+$$ and the following conditions:

1. $$f(x+2)\cdot f(x+3)=16$$,
2. $$f(x)+f(-x)=8$$,

we have to find integral $$\int_{-8}^8 f(x-2019)dx$$. So, if we plug in $$0$$, then $$f(0)+f(0)=8$$. Therefore, $$f(0)=4$$. Then plug in $$-2$$ in first equation, we get $$f(0)\cdot f(1)=16$$, therefore $$f(1)=4$$, analogically plug $$x=-1$$ and I think that $$f(x)=4$$, so answer will be $$64$$ (value after integrating). Is this correct? How can i prove it in a better manner?

• If $f$ was a polynomial, then $f \equiv 4$ would follow quite easily. If not, it can also be $f(x) = \sin(\pi x) + 4$, for example. Commented Dec 21, 2020 at 17:30
• I can prove that for integers, f(x)=4, but I am trying to prove for real numbers Commented Dec 21, 2020 at 17:33
• @mathboy Yes, but that does not seem right (at least I can't see yet). Look at the counterexample in my previous comment. Commented Dec 21, 2020 at 17:34
• @mathboy Are you sure that $f$ is not said to be polynomial? Commented Dec 21, 2020 at 17:38
• The solution to the functional equation is not unique, but the value of the integral is determined by the conditions. Commented Dec 21, 2020 at 18:28

## 2 Answers

From 1 we have $$f(x+1)f(x)=16$$, so $$\displaystyle f(x+2) = \frac{16}{f(x+1)} = \frac{16}{16/f(x)} = f(x)$$. Since $$f$$ has period $$2$$ then $$\int_{-8}^8 f(x-2019)\,dx = \int_{-8}^8 f(x-1)\,dx = 8 \int_0^2f(x-1)\,dx$$ $$= 8 \int_{-1}^1 f(x)\,dx = 8\int_{-1}^0f(x)\,dx+8\int_{0}^1 f(x)\,dx$$ $$=8 \int_{0}^1f(-x)\,dx + 8\int_{0}^1 f(x)\,dx = 8 \int_0^1f(x)+f(-x)\,dx = 8 \int_0^1 8\,dx = 64$$

• Interestingly, this solution is the same if we change $16$ by any other positive real. My second solution prove the stronger fact that $f$ is constant. That solution doesn't work if we change the $16$. Commented Dec 21, 2020 at 20:24

Here's a more direct solution:

From 1 we have $$\displaystyle f(1+x)=\frac{16}{f(x)}$$ and from 2 we have $$f(x) = 8-f(-x)$$.

Then $$f(x) = f(1+x-1) = \frac{16}{f(x-1)} = \frac{16}{8-f(1-x)} = \cfrac{16}{8-\cfrac{16}{f(-x)}} = \cfrac{16}{8-\cfrac{16}{8-f(x)}}$$

Solving for $$f(x)$$ we get $$f(x)=4$$.