$\lVert f_n \rVert_{\text{sup}} \to 0$ and $\frac{f_n(x)}{a_n} \to 0$ implies $\frac{\lVert f_n \rVert_{\text{sup}}}{a_n} \to 0 $?

Let $K$ be a compact subset of $\mathbb{R}^n$ and $f_n$ be a sequence of $C^1$ functions uniformly convergent to $0$ on $K$.

Also, let $a_n$ be a sequence of positive numbers converging to $0$ and assume that $\frac{f_n(x)}{a_n} \to 0$ for each $x \in K$.

Then, is it true that $\frac{\lVert f_n \rVert_{\text{sup}}}{a_n} \to 0 $ as well?

I am trying to prove or disprove this problem but it seems trickier than expected.. Could anyone help me?

Solution 1:

Hint: Take $K=[0,1]$. Draw a triangle with base $(\frac 1 n, \frac 2 n)$ and height $\frac 1 n$. Let $a_n=\frac 1 {n^{2}}$. That gives the graph of a continuous function such that for each $x$, $f_n(x)=0$ for $n$ sufficently large. For every $x$. So $\frac {f_n(x)} {a_n} \to 0$ but $\frac {\|f_n\|} {a_n} \to \infty$. These functions are not $C^{1}$ functions but you can modify them so as to make them smooth.

EDIT: Here is a modification that makes $f_n$'s belong to $C^{1}$: Take $f_n(x)=n^{3}(x-\frac 1n)^{2}(\frac 2 n -x)^{2}$ for $\frac 1 n \leq x \leq \frac 2n$ and $0$ elsewhere. [$a_n=\frac 1 {n^{2}}$ again].