Find a sequence of differentiable functions $f_n$ on $\mathbb{R}$ that converge uniformly to a differentiable function $f$, such that $f'_n$ converges pointwise but not uniformly to $f'$.


I have tried a number of possibilities, such as $f_n=x^n$ or $f_n=\frac{x^n}{n}$ but I don't know what the right approach is to construct the function. I am initially thinking that it's easiest to construct such a sequence of functions on the interval $[0,1]$ so that in the limit of $n$, part of the function goes to $0$ and the other part goes to $1$. However, this would make the resulting $f$ non-differentiable.


I think that$$f_n(x) = \dfrac{\sin(nx)}{n^2x}$$ will work. We have $f_n = \frac{1}{n}\textrm{sinc}(nx)$ and $\textrm{sinc}$ is uniformly bounded by $1$, so $f_n$ converges uniformly to zero.

Further, $$f_n' = \dfrac{\cos nx}{nx} - \dfrac{\sin nx}{n^2 x^2}$$ which should converge pointwise to zero, but $f_n'(\pi/n) = -1/\pi$ so the convergence can't be uniform. (These derivatives are all continuous too, with $f_n'(0) = 0$; you can convince yourself by writing out the Taylor series, $\cos y / y \sim 1/y - y/2 + O(y^3);$ $\sin y/y^2 \sim 1/y - y/6 + O(y^3)$ so the singularities "cancel"; alternately, you can use L'Hopital's rule).

(Here's a sketch that the derivatives converge pointwise to $0$: knowing that $f_n'(0) = 0$ for all $n$, just need to show that $$f_n' = \dfrac{1}{nx}(\cos nx - \textrm{sinc }nx)$$ can be made small for $x\neq 0$. The difference of cosine and sinc is bounded by $2$, and $\dfrac{1}{nx}$ can be made arbitrarily small for fixed $x \neq 0$.)

  • $\begingroup$ This is very interesting. I'm not familiar with the sinc function. Is your argument invalid of we replace sinc with $\sin$? $\endgroup$ – mathjacks Nov 26 '14 at 18:07
  • $\begingroup$ No; in that case the derivative is $\frac1n \cos nx$ which converges uniformly to $0$. Likewise, if you try $\frac1n \sin nx$, the derivative is $\cos nx$ which doesn't converge pointwise to anything. $\endgroup$ – BaronVT Nov 26 '14 at 19:12
  • $\begingroup$ You can read about the sinc function on wikipedia or whatever. The important thing to note is that $\lim_{y\to 0} \sin y / y = 1$ (which is easy to prove from scratch), so the function is continuous there. You can further show that the derivative is also continuous (as I've written above). $\endgroup$ – BaronVT Nov 26 '14 at 19:15

Take $f_{n}(x)=\sqrt{x^{2}+\frac{1}{n^{2}}}.$ Then for all $x\in %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ,$ $$\lim_{n\rightarrow \infty }f_{n}(x)=\left\vert x\right\vert =f(x).$$ So $% f_{n}$ converges pointwise to $f$ on $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion .$

Next, for any integer $n\geq 1,$ $$\left\vert f_{n}(x)-f(x)\right\vert =\frac{% 1}{n^{2}}\frac{1}{\sqrt{x^{2}+\frac{1}{n^{2}}}+\sqrt{x^{2}}}\leq \frac{1}{n}. $$ Then $f_{n}$ is uniformly convergent to $f$ over $% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion .$ Next, $f_{n}^{\prime }(x)=\frac{x}{\sqrt{x^{2}+\frac{1}{n^{2}}}}$ converges (pointwise) to $g(x)=\left\{ 0,\ if\ x=0,\ and\ x/\left\vert x\right\vert \ if\ x\neq 0.\right. $ Note that $f_{n}^{\prime }$ is continuous but not the limit $g$, so $f_{n}^{\prime }$ do not converges to $g $ uniformly.

  • 1
    $\begingroup$ My function f is not differentiable at $0$. $\endgroup$ – Idris Nov 26 '14 at 5:48
  • $\begingroup$ Yeah, there are at least a couple of "red herring" type non-examples. Besides sequences that don't have a differentiable uniform limit, I think you can also find examples where $f$ is differentiable, and $f_n'$ to some $g$ pointwise, but $g \neq f'$ (so it can't be said $f_n' \to f'$ pointwise, as required by OP). For example, $x/(1 + nx^2) \to 0$ uniformly (so $f' = 0$ as well), but the derivatives converge pointwise to $g = 1 \textrm{ if } x = 0; \ \ =0 \textrm{ if } x \neq 0$ $\endgroup$ – BaronVT Nov 26 '14 at 5:54

Let $f$ be a function which is everywhere zero except interval (0; 1) where it has some shape of a bell. Now the sequence is $f_n(x) = \frac{f(n(x - n))}n$
As you can see, "the bell" moves along the x-axis, becoming lower and lower, so the sequence uniformly converges to zero. But considering the derivative, we see that
$f'_n(x) = f'(nx - n^2)$, which always has maximum value equal to $sup f'(x)$. However, the location where it's non-zero also moves along the x-axis. So, point-wise it converges to zero, but not uniformly.
I think it's not too hard to construct a function with a "bell". For example, take $sin(x)$ on span [$-0.5\pi$; $1.5\pi$] and move it up.


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