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Find all $f\in C^1(\mathbb R,\mathbb R)$ such that $f^2+(1+f')^2\leq 1$

It's quite likely the answer is $f=0$.

Note that $|f|\leq 1$ and $-2\leq f'\leq 0$.

Therefore $f$ is decreasing and bounded.

What then ? I tried contradiction, without success.

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  • $\begingroup$ Have you thought about the limits as $x$ approaches $\pm \infty$? Remember that its values are decreasing and lie in a bounded region. $\endgroup$ – Kyle Jun 12 '14 at 18:53
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    $\begingroup$ well $-k\arctan(x)$ with $k$ small enough it seems for me satisfies your condition. Say $-1/\pi$. Then function is from $[-1/2,1/2]$ with negative derivative $>-2$. Something like that. $\endgroup$ – Alexander Vigodner Jun 12 '14 at 18:55
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    $\begingroup$ @AlexanderVigodner with $k=0.01$ it fails at $x=13$. $\endgroup$ – Gabriel Romon Jun 12 '14 at 19:17
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The equation is equivalent to $$ f^2+2f'+f'^2\le0\tag{1} $$ Since $f^2+2f'\le0$, where $f\ne0$, we have $$ (1/f)'\ge\color{#C00000}{1/2}\tag{2} $$ If $f(x_0)=a\gt0$, then $\dfrac1f(x_0)=\dfrac1a\gt0$ and $(2)$ says that $$ \frac1f\left(x_0-\frac3a\right)\le\frac1f(x_0)-\color{#C00000}{\frac12}\frac3a\lt0\tag{3} $$ as long as $\dfrac1f$ doesn't pass to $-\infty$ in $\left[x_0-\frac3a,x_0\right]$.

In any case, on $\left[x_0-\frac3a,x_0\right]$, $\dfrac1f$ must pass through $0$, which is impossible because $f\in C^1(\mathbb{R})$.

If $f(x_0)=a\lt0$, then $\dfrac1f(x_0)=\dfrac1a\lt0$ and $(2)$ says that $$ \frac1f\left(x_0-\frac3a\right)\ge\frac1f(x_0)-\color{#C00000}{\frac12}\frac3a\gt0\tag{4} $$ as long as $\dfrac1f$ doesn't pass to $\infty$ in $\left[x_0,x_0-\frac3a\right]$.

In any case, on $\left[x_0,x_0-\frac3a\right]$, $\dfrac1f$ must pass through $0$, which is impossible because $f\in C^1(\mathbb{R})$.

Therefore, $f(x)=0$ for all $x\in\mathbb{R}$.

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  • $\begingroup$ Can you elaborate a bit on your argument following $(2)$? $\endgroup$ – Alex Schiff Jun 13 '14 at 1:30
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    $\begingroup$ @AlexSchiff: Any function with a slope at least $\color{#C00000}{1/2}$ (e.g. $1/f$) must pass through the $x$-axis at a finite point. $\endgroup$ – robjohn Jun 13 '14 at 1:40
  • $\begingroup$ What is $f=0$ how you can use (2) then ? $\endgroup$ – Alexander Vigodner Jun 13 '14 at 5:18
  • $\begingroup$ @AlexanderVigodner: $f=0$ satisfies $(1)$. However, $(2)$ obviously holds wherever $f(x)\ne0$. $\endgroup$ – robjohn Jun 13 '14 at 6:12
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    $\begingroup$ @JackD'Aurizio: I have edited my answer to account for the case that $1/f$ may blow up inside an interval. $\endgroup$ – robjohn Jun 13 '14 at 14:28
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Since $f(x)$ is bounded and decreasing both $\lim_{x \rightarrow \infty} f(x)$ and $\lim_{x \rightarrow -\infty} f(x) $ exist. If $f(x)$ were not identically zero, then at least one of these limits is nonzero. Say it is the first one, and call the limit $L$.

By the mean value theorem, $f(n+1) - f(n) = f'(x_n)$ for some $x_n$ between $n$ and $n + 1$. The left-hand side of this equation converges to $L - L = 0$ as $n$ goes to infinity, so we have $$\lim_{n \rightarrow \infty} f'(x_n) = 0$$ But we also have $$\lim_{n \rightarrow \infty} f(x_n) = L$$ Plugging $x_n$ into $f(x)^2 + (1 + f'(x))^2 \leq 1$ and taking limits as $n$ goes to infinity gives $L^2 \leq 0$, a contradiction.

A similar argument works if $\lim_{x \rightarrow -\infty} f(x) \neq 0$.

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  • $\begingroup$ This is intuitive, the simplest argument so far. $\endgroup$ – Gabriel Romon Jun 13 '14 at 14:41
  • $\begingroup$ And this doesn't use the continuity of the derivative. Nice. $\endgroup$ – Gabriel Romon Jun 13 '14 at 14:46
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    $\begingroup$ You don't have to reason by contradiction, because you end up with $L^2 \leq 0 \implies L = 0$ anyway. $\endgroup$ – Najib Idrissi Jun 13 '14 at 14:55
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Hints: As you mentioned, $f$ is decreasing and bounded. Think about $\lim_{n \to \infty} f(n)$. Must this limit exist? What does this imply for the limit of the derivative $f'$?

Full Solution. The function $f(x)$ is decreasing and bounded, so $\lim_{x \to \infty} f(x)=L$ for some $L \in [-1,1]$. For the sake of contradiction, we suppose $|L|>0$. To set up the contradiction, we relate $|f(x)|$ and $f'(x)$: Let $\epsilon\in (0,1]$, and suppose that we have $0 \geq f'(x) \geq -\epsilon$ for some $x \in \mathbb{R}$. Then \begin{align*} f^2(x) & \leq 1-(1+f'(x))^2\\ &\leq -2f'(x) - (f'(x))^2 \\ &\leq -2f'(x) \\ & \leq 2\epsilon. \end{align*} Thus $|f(x)| \leq \sqrt{2\epsilon}$. Therefore we know that if $|f(x)| > \sqrt{2\epsilon}$, then $f'(x) <-\epsilon$. For sufficiently large $x$, we must have $|f(x)| > |L|/2=\sqrt{2(|L|^2/8)}$, hence $f'(x) <-|L|^2/8$. This contradicts the fact that $f(x)$ is bounded below. An entirely analogous argument shows that $\lim_{x \to -\infty} f(x)=0$. Monotonicity implies $f=0$.QED

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  • $\begingroup$ Although it should be a cakewalk, I fail to prove that for a $C^1$ function with a limit at infinity, its derivative must have $0$ as limit at $\infty$... $\endgroup$ – Gabriel Romon Jun 12 '14 at 19:05
  • $\begingroup$ Have you first shown that the limit of the derivative exists? $\endgroup$ – Kyle Jun 12 '14 at 19:11
  • $\begingroup$ I don't think you can prove the existence of the limit for the derivative with usual theorems. The only information about it is boundedness. $\endgroup$ – Gabriel Romon Jun 12 '14 at 19:14
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    $\begingroup$ And now that I remember this, math.stackexchange.com/questions/788813/… this is definitely wrong. $\endgroup$ – Gabriel Romon Jun 12 '14 at 19:19
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    $\begingroup$ Another thought would be to compare the limits as $x$ approaches positive and negative infinity. There are some easy restrictions there. $\endgroup$ – Kyle Jun 12 '14 at 19:25
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I think I have to completely rewrite the solution keeping the wrong one above untouched. As I said before I am sure we can build such function, and I think I did it using the ODE in the above solution in the end. I am building a counterexample function: Fistly, $f(x)= 0$ for $x\le 0$;

Now I'd like to build a simple function satisfying condition $$ f^2+(1+f^\prime)^2\le 1 $$ on some interval $[0,x^*]$.
I define $$ g(t)=-x^3/3-x^2/2\\ g^\prime(x)=-x^2-x $$ I is obvious that in some positive neighborhood $(0,\epsilon)$ $$ g^2+(1+g^\prime)^2 =O(\epsilon^4)+1-O(\epsilon) \le 1 $$ It is obvious also that staring from some $x^*$ $$ g^2+(1+g^\prime)^2\ge 1 $$ Besides we have this point $x^*$ is such that $$ g^2(x^*)+(1+g^\prime(x^*))^2= 1 $$ Let's check condition $g^\prime(x^*)>-1$. It is easy t estimate that $x^*$ is about $0.9$ and that then $g^\prime(x^*)< -1

Now again consider the differential equations $$ f^\prime=-1+\sqrt{1-f^2}\\ f^\prime=-1-\sqrt{1-f^2}\\ $$ and choose the second one in accordance with the sign $g^\prime(x^*)+1$. The solution of this equation with initial condition $f(x^*)=g(x^*)$ will extend our function on $R$. So the final function $f(x)$ is $$ f(x)=0 ~ if ~ x\le 0 \\ f(x)=g(x) ~ if~ 0<x\le x^* \\ solution~ of~ f^\prime=-1-\sqrt{1-f^2}, f(x^*)=g(x^*) ,~ x\ge x*\\ $$

Since I did not find any mistake in the proof that such function cannot exist I again probably made mistake somewhere. But I cannot find it. Any comments please. May be this ODE cannot have a solution on $R$? Lipschitz condition is not satisfied.

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FIX Well the solution of differential equation with non zero initial condition will satisfy your property $$ f'=-1-\sqrt{1-f^2} $$ Edit Notice that equation $$ f'=-1+\sqrt{1-f^2} $$ is also OK. Now let's change $\tau = -t$ and rewrite the second equation as function of $\tau$ $$ f_\tau^\prime =1-\sqrt{1-f^2} $$ Let's build now the function on $R^+$ as a solution of the first eqaution and on $R^-$ as a solution of the third equation. To guarantee differentiability in $t=0$ let's make equal derivatives at time $t=\tau=0$. $$ f^\prime_t=-1-\sqrt{1-f^2}=-f^\prime_\tau=-1+\sqrt{1-f^2} $$ So with initial condition $f=1$ we can propagate this function on $R$. I changed signs here. Now the initial derivative is

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  • $\begingroup$ Can you guarantee that it will exist on all of $\mathbb{R}$ if $f$ is not $\equiv 0$? $\endgroup$ – Daniel Fischer Jun 12 '14 at 19:24
  • $\begingroup$ I think we can play a bit with this equation to guarantee this. Let me think $\endgroup$ – Alexander Vigodner Jun 12 '14 at 19:27
  • $\begingroup$ You can't have $f(x) = 1$ for any $x$. That would imply $f'(x) = -1$, and thus $f(y) > 1$ for $x-\varepsilon < y < x$. $\endgroup$ – Daniel Fischer Jun 12 '14 at 19:41
  • $\begingroup$ Who said this. I said it is 1 in the inital point. After this you extend it in both direction via equations. Extention is built on the different equation it is not symmetric. $\endgroup$ – Alexander Vigodner Jun 12 '14 at 19:43
  • $\begingroup$ OK. I am wrong about negative direction. But something can be done, I am sure. I don't belieive f=0. $\endgroup$ – Alexander Vigodner Jun 12 '14 at 19:47

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