# Do the property of $f_n$ carry over to $f_n'$?

Suppose a sequence of differentiable functions $$f _ { n } : \mathbb { R } \rightarrow [ 0,1 ]$$ converges pointwise to the zero function. Does it follow that the derivatives $$f _ { n } ^ { \prime }$$ converge pointwise to the zero function?

No! Consider $$f_n(x)=\frac{\sin nx}{\sqrt{n}}$$ on $$\Bbb R$$. Here each $$f_n$$ is infinitely differentiable and $$f_n$$ converges pointwise (actually uniformly) to zero. But $$f_n'(x)=\sqrt{n} \cos nx$$ is divergent at each $$x \in \Bbb R$$
• remember, you make the pointwise convergence as zero. Also you know $1/n, 1/n^2,...,1/ \sqrt{n}$ are all converges to zero. so construct $f_n$ such as $$f_n(x)=\frac{?}{n}\;\;\text{or}\;\;\frac{?}{\sqrt{n}}\;\;\text{or}\;\;\frac{?}{n^2}\;\text{or}\;\;\frac{?}{\sqrt{1+n}} \cdots$$ Also make numerator as differentiable and bounded with its derivative is not differentiable – Chinnapparaj R Dec 19 '18 at 3:16
• For example, $$f_n(x)=\frac{nx}{1+n^2x^2}$$ is another counterexample to your statement! – Chinnapparaj R Dec 19 '18 at 3:20
• You can intuit your way to other functions that work with this method as well - e.g. another is $f_n(x) := \frac{1}{\sqrt{n}} e^{-[2^n (x - 0.5)]^2}$, based on the idea of a gaussian as a "nascent" delta function and ignoring the normalization as irrelevant for this case. The peak in the middle gets shorter but gets sharper much faster, so the derivatives on its sides pop out like bulging eyeezzzzz...... hggeeeeeggggghhhhieeeeee....... – The_Sympathizer Dec 19 '18 at 4:35