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Exercise: Find all of functions $f: \mathbb{R} \to \mathbb{R}$ such that $$f(x+y^2 + z^3) = f(x) + f^2(y) + f^3(z), \forall x,y,z \in \mathbb{R}.$$

In fact, I solved this question as follows:

  • First step: Calculate $f(0)$. In this step, I obtained that $f(0) = 0$ or $f(0) = - 1$.

  • Second step: Consider two cases of the first step. I determinated the solutions are $f(x) = 0$, $f(x) = -1$ and $f(x) = x$.

The problem is that this way is very long. Who has another way to solve my exercise? Thank you very much for your interests.

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    $\begingroup$ What was your solution that was "very long"? We can't be sure that we aren't just reproducing whatever solution you found. $\endgroup$ – Tob Ernack Jul 19 '17 at 5:27
  • $\begingroup$ @TobErnack Thank you so much. In fact, I'd like a method which is shorter than my way. $\endgroup$ – quoc hoan Ngo Jul 19 '17 at 6:15
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Well this is not the most elegant of proofs either...

Case where $f(0) = 0$

We see that $f(x + y) = f(x + 0^2 + (\sqrt[3]y)^3) = f(x) + f(0)^2 + f(\sqrt[3]y)^3 = f(x) + f(y)$.

We also see that if $x \leq y$ then $f(x) \leq f(y)$ because $f(y) = f(x + y - x) = f(x + \sqrt{y - x}^2 + 0^3) = f(x) + f(\sqrt{y - x})^2 + f(0)^3 \geq f(x)$.

From the first relation above, we find that $f\left(\frac{1}{n}\right) = \frac{f(1)}{n}$ for $n \in \mathbb{N}^*$.

For any $\epsilon \gt 0$, choose $n$ large enough so that $\frac{|f(1)|}{n} \lt \epsilon$. Choose $\delta = \frac{1}{n}$.

Then whenever $|x| \lt \delta$, we find that $|f(x)| \lt |f(\delta)| = |f(1/n)| = \frac{|f(1)|}{n} \lt \epsilon$ which shows that $\lim\limits_{x \rightarrow 0} f(x) = f(0) = 0$.

Finally we see that $\lim\limits_{h \rightarrow 0}f(x + h) - f(x) = \lim\limits_{h \rightarrow 0}f(h) = 0$ so $f$ is continuous everywhere.

$\frac{f(x)}{x}=\frac{f(rx)}{rx}$ for all $r \in \mathbb{Q}$ shows that $f(x)/x$ is constant on a dense subset of $\mathbb{R^+}$, and by continuity and even-ness, must be constant everywhere on $\mathbb{R}^*$.

So on $\mathbb{R}^*$ we have $f(x) = Ax$ for some $A$ which can be found to satisfy $-A = A(-1 + (-1)^2 + (-1)^3) = -A + A^2 - A^3$ or $A \in \{0, 1\}$.

So $f(x) = x$ or $f(x) = 0$ on $\mathbb{R}^*$, and thus also on all of $\mathbb{R}$.


Not yet worked out the case where $f(0) = -1$.

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  • $\begingroup$ Oh, your method is nice. Thank you @Tob Ernack $\endgroup$ – quoc hoan Ngo Jul 19 '17 at 6:22
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The case $ f ( 0 ) = 0 $ is solved by @TobEnrack. For the case $ f ( 0 ) = -1 $, let $ x = y = 0 $ in $$ f \big( x + y ^ 2 + z ^ 3 \big) = f ( x ) + f ( y ) ^ 2 + f ( z ) ^ 3 \tag { 0 } $$ and you get $$ f \big( z ^ 3 \big) = f ( z ) ^ 3 \tag { 1 } $$ Again, letting $ x = z = 0 $ in $ ( 0 ) $ you have $$ f \big( y ^ 2 \big) = f ( y ) ^ 2 - 2 \tag { 2 } $$ Now combining $ ( 1 ) $ and $ ( 2 ) $ you get $$ f ( x ) ^ 6 - 2 = \big( f ( x ) ^ 3 \big) ^ 2 - 2 = f \big( x ^ 3 \big) ^ 2 - 2 = f \Big( \big( x ^ 3 \big) ^ 2 \Big) = f \big( x ^ 6 \big) \\ = f \Big( \big( x ^ 2 \big) ^ 3 \Big) = f \big( x ^ 2 \big) ^ 3 = \big( f ( x ) ^ 2 - 2 \big) ^ 3 = f ( x ) ^ 6 - 6 f ( x ) ^ 4 + 12 f ( x ) ^ 2 - 8 $$ By simple algebra, it's easy to derive $$ f ( x ) ^ 2 = 1 \tag { 3 } $$ Finally, letting $ z = 0 $ in $ ( 0 ) $ and using $ ( 3 ) $ you get $$ f \big( x + y ^ 2 \big) = f ( x ) \tag { 4 } $$ Hence, if $ x \le y $, substituting $ \sqrt { y - x } $ for y in $ ( 4 ) $, it's easy to see that $$ f ( y ) = f \big( x + ( y - x ) \big) = f \Big( x + { \sqrt { y - x } } ^ 2 \Big) = f ( x ) $$ Since for every $ x $, either $ x \le 0 $ or $ 0 \le x $, therefore $ f ( x ) = -1 $.

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