I'm doing an exercise from baby Rudin (chapter 8 exercise 11) and found a suggestion that it might use.

$$\left|\int_0^\infty e^{-x}f\left(\frac{x}{t}\right)dx\ -1\right| \leq \int_0^\infty e^{-x}\left|f\left(\frac{x}{t}\right) -1\right|dx$$

However, I don't quite understand what rule or theorem makes this possible. I think it might have something to do with the integral of $e^{-x}$ from 0 to infinity being 1 but am not sure. Does anyone have insight or suggestions as to what I might investigate to get my head around it?

  • 3
    $\begingroup$ Replace the first $1$ with $\int_0^{\infty}{e^{-x}\,dx}$. $\endgroup$ – Mindlack Feb 12 at 22:26
  • $\begingroup$ ahhhh ok that helps a lot. Thank you for the tip. I'll write it out and post if that gets me to understand it. brb $\endgroup$ – Algebra is Awesome Feb 12 at 22:27
  • $\begingroup$ Ok, I checked it out and I believe I understand the inequality now. Thank you very much! $\endgroup$ – Algebra is Awesome Feb 12 at 23:00

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