4

Let $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ be a function satisfying $$f(x)f\big(yf(x)\big)=f(x+y),\forall x,y\in\mathbb{R}^+$$ If $f(1)=\frac{1}{152}$, evaluate $f(4)$.

By inspection, we can see $f(x)=\frac{1}{151x+1}$ is a solution, from which we can easily get the answer. But how can we show that this is the only solution?

Here is my work:

Because $f$ is the reciprocal of a linear function, it would probably help to define $g(x)=\frac{1}{f(x)}$ (note that this is well defined as we are working in $\mathbb{R}^+$). Then the given equation becomes $$g(x)g\left(\frac{y}{g(x)}\right)=g(x+y)$$

If we take $y=g(x)$, then this becomes $$g(x)g(1)=g\big(x+g(x)\big)\implies 152g(x)=g\big(x+g(x)\big)$$

Not sure where to go from here. Any thoughts?

Mohsen Shahriari
  • 6,718
Max
  • 1,408
  • Do possible solutions to the functional equation have to be continuous? – stressed out Nov 24 '17 at 02:21
  • Are you sure you got computations right? Because the equation does not holds for $x=y=1$ with the claimed solution. – mucciolo Nov 24 '17 at 02:23
  • Whoops, that was a typo. Thanks for the catch. – Max Nov 24 '17 at 02:25
  • I am not sure how taking the reciprocal of $f$ would help. Try exploiting stronger properties of the solution you got, such as differentiability and invertibility. On a side note, the denominator of $f$ is not a linear but an affine function. – mucciolo Nov 24 '17 at 02:47
  • My motivation for taking the reciprocal was that proving the only solution to $g$ is $151x+1$, a linear function, is easier than proving the only solution to $f$ is $\frac{1}{151x+1}$. I was trying to do this through reducing the equation to Cauchy’s Functional Equation, which, under certain conditions, has solutions $f(x)=cx$. – Max Nov 24 '17 at 04:10
  • Also, this is the problem statement as is. There is no mention of continuity, though now I think maybe it’s necessary to prove the only solution is the one I mentioned. Can anyone prove me wrong? – Max Nov 24 '17 at 04:11
  • 1
    On second thought, the question did not ask to find all $f$, it merely asked to find $f(4)$. Maybe it’s possible to do this without proving uniqueness of the above solution? – Max Nov 24 '17 at 04:36
  • It is very likely that there is a clever way of finding only $f(4)$ because assigning a value to $f(1)$ is enough to guarantee solution uniqueness. (On a side note, if you want to interact with other users in the comments section of your posts you need to tag them because so we will get notified of your responses to our comments. Otherwise we are not notified.) – mucciolo Nov 24 '17 at 20:41
  • related: https://math.stackexchange.com/questions/2300513/solving-the-functional-equation-f-big-x-y-f-x-big-f-x-f-y – Mohsen Shahriari Nov 26 '17 at 08:16
  • @MohsenShahriari Interesting. Looks like taking $\mathbb{R}$ as the codomain and range gives some interesting pathological solutions. – Max Nov 26 '17 at 16:39
  • Does this answer your question? To find all functions $f: (0,\infty) \rightarrow (0, \infty)$ such that $f(x+y) = f(x)f(yf(x))$ – Mohsen Shahriari Sep 30 '21 at 22:16

1 Answers1

4

I'm going to proceed with your $g(x) = 1/f(x)$, starting with this $$g(x)g\left(\frac{y}{g(x)}\right)=g(x+y)$$ but then re-setting in it $y = g(x)y_1$ (possible for any positive $y_1$) to get this basic equation $$\tag1 g(x)g(y)=g(x+g(x)y)$$ (in which I renamed $y_1$ back to $y$). Applying this first with $x=1$ and then with some other $x = w > 1$, letting $B := g(w)$, gives $$ g(1+152y) = 152g(y)\\ g(w+By) = Bg(y) $$ Suppose $B < 152$. We can then pick a positive $y$ such that $1+152y = w + By$ so that the left-hand sides are equal, but the right-hand sides are not (since $g(y) > 0$) - contradiction. So $w > 1$ implies $g(w) \geq 152$.

Now consider $u = w + B(1+152y_0)$ and $v = 1+152(w+By_0)$ for some arbitrary $y_0 > 0$: $$ g(u) = g(w + B + 152By_0) = Bg(1+152y_0) = 152Bg(y_0)\\ g(v) = g(1+ 152w +152By_0) = 152g(w+By_0) = 152Bg(y_0) $$ Since the right-hand-sides are equal, $g(u) = g(v)$. This means that either $w+B = 1+152w$, meaning that $$\tag2 g(w) = B = 1+151w $$ as we wanted to prove, or, $g(u) = g(v)$ with $u \neq v$. We derive a contradiction in the latter case. WLOG, suppose $u < v$. Then setting in (1) $x = u, y = (v-u)/g(u) =: t$, we get $$ g(u)g(t) = g(u + g(u)t) = g(v)\\ \therefore g(t) = 1 $$ Now set in (1) $x = t$, to get $g(t+y) = g(y)$ and since $g(t) = 1$, it follows by induction that $g(nt) = 1$ for any positive integer $n$. By choosing a sufficiently large $n$, we can make $nt > 1, g(nt) = 1$ which is a contradiction with the previous result that $w > 1 \Rightarrow g(w) \geq 152$. This leaves (2) as the unique solution.

Nick Pavlov
  • 1,553
  • Awesome solution! If you don't mind me asking, what was the motivation behind it? – Max Nov 24 '17 at 17:11
  • 2
    Of course I don't mind - those are the best questions. I was playing around with that equation (1), trying to get some sort of recursion going (I didn't really think I can prove uniqueness, was just trying to evaluate g(4)), and when I wrote it at first just for $x=1$ and $x=4$ I noticed how I can bring them together - sort of like in solving simultaneous equations by elimination. Before that I had tried to prove that it has to be injective, which played into the second part. So, really, no magic answer, just writing down everything that came to mind and fiddling with it.@Max – Nick Pavlov Nov 24 '17 at 17:19