Prove $$\lim_{|k|\to \infty}{\mathscr F f(k)}=\lim_{|k|\to \infty} \int_{-\infty}^\infty f(x)e^{-ikx}dx=0$$ (imagine the low bound is negative infinity).
I tried integrating by parts... where the first term is evaluated at x from -infinity to +infinity $$\lim_{|k|\to \infty} {\left(\frac{-1}{ik}\right)f(x)e^{-ikx} - \int_{-\infty}^\infty \left(\frac{-1}{ik}\right)f'(x)e^{-ikx} dx}$$
This second integral can be substituted for the original integral (imagine the low bound is negative infinity), $$\int_{-\infty}^\infty f(x)e^{-ikx}dx$$ by properties of Fourier Transforms, which would allow a cancellation however I still cannot prove the first part (the uv part of integration by parts) will go to zero.
The hint is to assume that since f'(x) is absolutely integrable that $$\lim_{|x|\to \infty} f'(x) = 0$$, however I do not see how I can do this when f'(x) will always be inside of the integrand.
The next step is to broaden this argument to show that the $$\lim_{|k|\to\infty}{|k|^{n-1} \mathscr F f(k)=0}$$ as k approaches infinity of |k|^(n-1)Ff(k)=0 when f is n times differentiable and the jth derivative of f is absolutely integrable where j=1,...,n. Any help is much appreciated!
Actually the first term evaluated from - infinity to infinity goes to zero without application of the limit since f(x) is absolutely integrable and e^ix is bounded. After that the integral is bounded since f'(x) is absolutely integrable and e^ix is bounded again. Since 1/k will go to zero, the entire limit is 0