# What do the frequency components of a fourier transform do to the function itself?

I thought for a while intuitively about the relationship between the frequencies in the fourier transform of a function and the function itself.

After a while, I figured out that higher frequencies cause a function to be "jumpier" and less smooth. Is this interpretation true?

Do the frequencies in the fourier representation of a function represent the "smoothness" of the function in time domain? This would explain why many functions, when you cutoff their high frequency components, become smoother.

• You may want to check out the Riemann Lebesgue Lemma and its proof. Basically, you use integration by parts to translate statements about decay of fourier coefficients to statements about $f$'s derivatives. – SquirtleSquad Mar 17 '17 at 1:33
• Here's an example of the technique I described: math.stackexchange.com/questions/2090803/… – SquirtleSquad Mar 17 '17 at 1:36

If all goes well, the Fourier transform of the derivative is the derivative of the transform. But note that the high frequency bits become much bigger when you take derivatives. Therefore the low frequency bits encode slowly changing regions of the signal, while the high frequency bits encode the fastest changing portions.

If you have a sharp boundary anywhere, it is almost entirely going to be in the high frequency bits. Remove those and you remove the artifact. (And get a certain amount of "ringing" throughout the entire signal...to improve on that you should look at wavelets instead of sin/cos series.)