I'm trying to find a smooth function $f: \mathbb{R} \times (0, \infty) \rightarrow \mathbb{R}$ which satisfies the following conditions:

1) $\lim_{t \rightarrow 0^+} t\cdot \left(\frac{\partial f}{\partial x}\right)^2$ exists and is not identically $0$ as a function of $x$.

2) $\lim_{t \rightarrow 0^+}t\cdot \left(\frac{\partial f}{\partial x}\right)\left(\frac{\partial f}{\partial t}\right)$ exists for all $x$.

3) $\lim_{t \rightarrow 0^+} t\cdot \left(\frac{\partial f}{\partial t}\right)^2$ exists for all $x$.

I've proved that $f$ can't be separable (i.e. $f(x,t)$ can't be of the form $f(x,t) = X(x)T(t)$). In fact I'm starting to believe that no such function can exist, but I haven't been able to prove it. Any help would be greatly appreciated!

  • $\begingroup$ How do you prove that the function can not be separable? $\endgroup$ Jan 25, 2016 at 19:40
  • $\begingroup$ @DanielRobert-Nicoud I think 1 and 2 can be combined to show that f can't be separable, since otherwise from 1 $T(t)$ show contained $1/\sqrt t$, which will not satisfies 2. $\endgroup$
    – k99731
    Jan 25, 2016 at 19:50
  • $\begingroup$ @k99731 Right, thanks. I did the computation but didn't realize... $\endgroup$ Jan 25, 2016 at 19:51

1 Answer 1


A (very) partial solution: From the first condition, we know that $\partial_xf(x,t)$ must be composed by a part that goes like $t^{-\frac{1}{2}}\tilde{g}(x)$ as $t\to0^+$ (where $\tilde{g}$ is some smooth function not identically zero), plus some other function $\tilde{h}(x,t)$ that satisfies $$\lim_{t\to0^+}\left(2\sqrt{t}\tilde{g}(x)\tilde{h}(x,t) +\tilde{h}^2(x,t)\right) = 0.$$ It follows that $\tilde{h}\to0$ as $t\to0^+$.

Knowing this, let's work in the simplified assumption that $$f(x,t) = \frac{g(x)}{\sqrt{t}} + h(x,t),$$ where $g$ is smooth and $h$ is analytic in $t$ with no constant term. The first condition is automatically satisfied. The third condition asks for the limit of \begin{align} t(\partial_tf)^2 = & t\left(-\frac{1}{2}\frac{g}{(\sqrt{t})^3} + \partial_th\right)\\ = & \frac{1}{4}\frac{g^2}{t^2} -\frac{g\partial_th}{\sqrt{t}} + t(\partial_th)^2 \end{align} as $t\to0^+$ to exist. The last term $t(\partial_th)^2$ gives zero in the limit, and noticing that $\partial_th$ goes to a constant $k\in\mathbb{R}$ we get $$\lim_{t\to0^+}\frac{1}{4}\frac{g^2}{t^2} -k\frac{g}{\sqrt{t}}.$$ This cannot converge, so such a solution is impossible.

  • $\begingroup$ Okay cool. How about we define $h(x,t)$ by $f(x,t) - g(x)/\sqrt{t}$, so that property 1 is satisfied. Then does demanding properties 2 and 3 on $h$ contradiction property 1? I'll look into it. Thanks. $\endgroup$
    – Eric Ling
    Jan 25, 2016 at 21:12
  • $\begingroup$ @EricLing I'm not sure... I used the fact that $h$ was assumed to be analytic in my reasoning, but we know that there are very nice functions that are not... $\endgroup$ Jan 25, 2016 at 21:35
  • $\begingroup$ So it seems like if a solution exists, it won't be something that we can nicely write down. $\endgroup$
    – Eric Ling
    Jan 25, 2016 at 22:10
  • $\begingroup$ @EricLing Not necessarily. For example the function $f(x) = e^{-\frac{1}{x^2}}$ for $x>0$ and $0$ for $x\le 0$ is smooth but not analytic. $\endgroup$ Jan 25, 2016 at 23:29
  • $\begingroup$ Ah yes that's right. Good point. $\endgroup$
    – Eric Ling
    Jan 26, 2016 at 2:03

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