1

Let $f:[-\pi, \pi] \rightarrow \mathbb{C}$ be a Riemann-integrable function that is continuous at zero.

Since $f$ is continuous at $0$, we can choose $0 \lt \delta \leq \pi/2$, so that $f(\theta) \gt f(0)/2$ whenever $|\theta| \lt \delta$.

I'm not seeing how they got that from continuity.

Daniel Donnelly
  • 21,342

1 Answers1

5

Hint: In the definition of continuity, choose $\epsilon = f(0)/2$, then there exists $\delta$ so small that if $|\theta| < \delta$ then $|f(0) - f(\theta)| < f(0)/2$. Your statement follows from this.

treble
  • 4,094
  • 18
  • 21
  • The part I'm not getting is how you get $f(\theta) \gt f(0)/2$ from that. – Daniel Donnelly Sep 20 '13 at 23:47
  • I think I got it. If $f(\theta) \leq f(0)/2$, then $f(0) - f(\theta) \geq f(0)/2$ Taking absolute value (and $f(0) \gt 0$ given), gives a contradiction so $f(\theta) \gt f(0)/2$. QED – Daniel Donnelly Sep 21 '13 at 00:08