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The problem is as follows (verbatim) : Prove that it is impossible to write $x = f(x)g(x)$ where $f$ and $g$ are differentiable and $f(0)=g(0)=0$.

I rephrased the question as I realized it wasn't optimally worded : Prove that it is impossible to write $x = f(x)g(x)$ for all $x \in \mathbb R$, where $f$ are $g$ are differentiable on $\mathbb R$ and $f(0)=g(0)=0$.

I proved this by differentiating both sides of $x = f(x)g(x)$, which gives $1 = f'(x)g(x) + f(x)g'(x)$; thus when $x=0, f'(0)g(0) + f(0)g'(0) = 0 \neq 1$. However, this proof doesn't give me much insight on why this particular exercise is true. I am really interested in seeing a more intuitive explanation, but here are some of my attempts/hypotheses :

  1. $f'(x)g(x) + f(x)g'(x)$ is constant when $f$ or $g$ is constant. However, WLOG, if $f(x) = 0$, then we will have to divide by zero in order to get a function $g$ such that $x = f(x)g(x)$.
  2. If we define $f(x) = \frac{x}{g(x)},$ then clearly if $g(x) = 0$ then $f$ is not even continuous at $x=0$, so $f$ would not be differentiable on $\mathbb R$.

I am more or less looking for a graphical(?), intuitive explanation. Thank you for your help!

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  • $\begingroup$ i upvoted. nice analysis. interesting question. i don't know the answer either. $\endgroup$ – user2661923 Mar 31 at 2:25
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    $\begingroup$ This might be a possible explanation. Since $f$ and $g$ are differentiable, they can be approximated by linear functions near $x = 0$. Hence $f(x) g(x)$ looks quadratic near $0$ and vanishes faster than $x$ does. $\endgroup$ – SFeesh Mar 31 at 2:57
  • $\begingroup$ Your attempt 2. doesn't really highlight any issues. If you say $f(x)=\frac{x}{g(x)}$ for $x\neq 0$, it's possible that $x\mapsto \frac{x}{g(x)}$ has a hole at $x=0$. In this case, you can fill in $f(0)=\lim_{x\to 0} \frac{x}{g(x)}$. $\endgroup$ – zugzug Mar 31 at 4:07
  • $\begingroup$ Even if there is a removable discontinuity there, I don't think $f(x) = \frac{x}{g(x)}$ will be differentiable anyway... right? $\endgroup$ – alexion Mar 31 at 5:13
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We have by given conditions $f(x) /x\to f'(0),g(x)/x\to g'(0)$ as $x\to 0$. Hence on multiplication we have $$\frac{f(x) g(x)} {x^2}\to f'(0)g'(0)$$ or $$\frac{1}{x}\to f'(0)g'(0)$$ which is an obvious contradiction.


You can translate the above argument into a more informal language. Differentiability of $f, g$ at $0$ ensures that they behave like linear functions in neighborhood of $0$ and both vanish at $0$ so we must have $$f(x) \approx ax, g(x) \approx bx$$ in some neighborhood of $0$ where $a, b$ are constants. But then $$x=f(x) g(x) \approx abx^2$$ or $$1\approx abx$$ which is absurd as right hand side can be made small by choosing $x$ small but left hand side is constant.

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    $\begingroup$ Very nice! I think even if I came up with the intuitive idea, my rigorous implementation would've been much messier. Do you have any suggestions for remembering and applying this idea? $\endgroup$ – Ovi Apr 1 at 4:49
  • $\begingroup$ @Ovi: I think much of my intuition comes from a deep/rigorous understanding of basic real analysis (thanks to Hardy for writing A Course of Pure Mathematics) and overall experience dealing with many problems in this area (thanks to MSE). $\endgroup$ – Paramanand Singh Apr 1 at 7:06
  • $\begingroup$ Ah okay thanks! $\endgroup$ – Ovi Apr 1 at 16:23
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Here's a pretty picture. Let $h(x)=f(x)g(x)$ and assume $h$ is continuously differentiable. Then $h'(0)=0$ so by continuity, for $\epsilon>0$ small, there exists a $\delta>0$ such that when $x\in [-\delta,\delta]$, we have $|h'(x)|\leq\epsilon$. In other words, for $x_1,x_2 \in [-\delta,\delta]$, by the mean value theorem, $|h(x_2)-h(x_1)|\leq \max_{[x_1,x_2]}|h'(x)|\, |x_2-x_1|\leq \epsilon\, |x_2-x_1|$ so $h$ is locally a contraction mapping.

Moreover, for $x\in[-\delta,\delta]$, we have $$ |h(x)|=|h(x)-h(0)|\leq\epsilon|x|\leq \epsilon\delta\leq\delta $$ so $h$ maps $[-\delta,\delta]$ to itself.

In conclusion, both criteria to apply Banach's fixed point theorem are met. It follows that $h$ has a unique fixed point, namely $h(0)=0$.

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  • $\begingroup$ That is a pretty picture, but I'm not quite sure it gets at the heart of the matter. I think the OP was looking for a way to see why $h'(0) = 0$, as they derive it, but don't understand the why. Here you seem to state it as a trivial consequence of the product rule. $\endgroup$ – Isaac Browne Apr 1 at 3:41
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I believe that the problem here is a lack of intuition for the product rule. You apply it to derive $h'(0) = 0$, but you seem to not see intuitively why it must be $0$ and not $1$.

3Blue1Brown has a nice exposition of the general intuition behind the product rule (and the chain rule!) here.

Thus, instead of trying to replicate that, I'll just speak about how to apply it here.

Most people will show you the following picture for the product rule:enter image description here

The strange thing about this case is that the part that matters to us is "Area C". This is because $x=0$, so $f(x)=g(x)=0$. Thus this will be the only nonzero area in the entire picture!

Now, the other critical piece of intuition is to observe how differentiability implies that this Area C goes to $0$ too fast to be included in the derivative of $f(x)g(x)$. To see this, I recommend Paramanand Singh's wonderful answer to this very question.

I especially recommended the latter part of that answer, on how the derivatives are basically good local linear approximations $ax,bx$. Then if we multiply two of those together, we get a quadratic $abx^2$, which at it's extreme point, has slope $0$.

Hope that helps!

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    $\begingroup$ I am glad you liked my answer. +1 for images and yes that explains the product rule nicely. $\endgroup$ – Paramanand Singh Apr 1 at 7:11
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I think the fact that f and g are differentiable is important not because you need to find the derivative of f(x)g(x), but because this means that the two functions are continuous and will not have a cusp or asymptote. This means that rational functions and logarithmic functions are not a possibility - which will be important.

For f(x)g(x) to result in x for all x on R, it must be the case the f and g cancel each other out and leave only x behind. This is not possible if f and g are both polynomials as their product will increase in degree and will not result in x. Trigonometric functions will either have asymptotes or a limited range. Finding a way to prove that there is no pair of functions that will be both differentiable and cancel each other out cleanly will be the way forward.

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