0

Let $I=[a,b]$ be an interval and $x_0 \in I$ fixed. I'm trying to show that for any $\epsilon>0$ exists a continuously differentiable function $f$ such that $|f'(x_0)| \geq \epsilon \| f \|_{\infty}$.

Can someone give me a hint on how to construct such function? I'm obviously missing something here.

blomp
  • 494
Mathbds
  • 83
  • Think of a differentiable function that is increasing such that $f(a)=0$ and $f(b)=1$. Then just alter the slope of $f$ at $x_0$ to be as steep as you need – JackT Feb 27 '24 at 22:51
  • Hint: Think of variations of the sequence of functions $f_n(x)=\sin(nx)$. The supremum norm of such functions is bounded by $1$, but the derivative might be very large. – Mark Feb 27 '24 at 22:55

0 Answers0