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QUESTION : Find all functions $f:\mathbb{R} \mapsto \mathbb{R}$ which satisfy $f(x^2+y f(z)) =x f(x) + z f(y)$

My doubt lies in the part where I've shown injectivity. Kindly check if my proof is correct.

The solution is $f(x)=x$ for every individual $x$. Let $P(x,y,z)$ be an assertion of the FE.

  • $P(0,0,0)\equiv f(0)=0$
  • $P(x,0,0)\equiv f(x^2)=xf(x)$
  • $P(0,y,z)\equiv f(yf(z))=zf(y)$

Does the third condition imply injection in the following manner?

If $f(a)=f(b)$, then $f(yf(a))=f(yf(b))\implies af(y)=bf(y)\implies a=b$ assuming $f(y)\neq 0 \forall y$.

If till this it's correct, I am pretty much done with the whole problem.

Here's how I can proceed henceforth— $f(x^2)=xf(x)=f(xf(x))$, which, upon using the injectivity criteria leads to $x^2=xf(x)$. Now, $f(x)$ can be found after verification of the positive/negative value.

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your solution is almost correct - you arrived at $f(a)=f(b)\Rightarrow{}af(y)=bf(y)$ for all $y\in{}\mathbb{R}$. Now you assumed that $f(y)\ne{}0$ for all $y$ (which solves the question but isn't justified). Instead it is enough to assume that $f(y)\ne{}0$ for some $y$. Note that if this does not hold, then $f(x)=0$ for all $x$, which is another solution.

Also, $xf(x)=x^2$ implies immidiatly that $f(x)=x$ for all $x\ne{}0$ without further checking. Did you prove that $f(0)=0$ though?

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You only need to assume that there is some $y$ with $f(y)\neq 0$ in order for your argument on injectivity to hold. In this case your work shows that $f(x)=x$ for all $x\neq 0$, and since $f$ must be injective, the only possibility for $f(0)$ is $f(0)=0$. This concludes the proof that $f(x)=x$ for this case.

Now, the other case is simple, and ties in with Cameron's comment that $f\equiv 0$ is another -- and hence the only other -- solution.

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