@BalarkaSen Go to Tibet
Ride a camel.
Read the bible.
Dye your shoes blue.
Grow a beard.
Circle the world in a paper canoe.
Subscribe to The Saturday Evening Post.
Chew on the left side of your mouth only.
Marry a woman with one leg and shave with a straight razor.
And carve your name in her arm.

Brush your teeth with gasoline.
Sleep all day and climb trees at night.
Be a monk and drink buckshot and beer.
Hold your head under water and play the violin.
Do a belly dance before pink candles.
Kill your dog.

Let $f,f'$ piecewise continuous functions in $-L<x<L$ and $f$ 2L-periodic, then I want to show that $na_n$ and $nb_n$ are bounded sequences, where $a_n, b_n$ Fourier coefficients of $f$.

We have that f is piecewise continuous, so there are subintervals of [0,L], [-L,0] at which f is bounded and $f(x)=\frac{a_0}{2}+ \sum_{n=1}^{\infty} \left( a_n \cos{\frac{n \pi x}{L}}+ b_n \sin{\frac{n \pi x}{L}}\right)$.

Since the functions $\sin{x}, \cos{x}$ are bounded, in order f to be bounded, $a_n$ and $b_n$ need to be bounded, and so $na_n$ and $nb_n$ also are.

@TedShifrin $1 \leq 1$ but $n \cdot 1 \to +\infty$, so it is not bounded. @TedShifrin

I have thought the following.

We can get that $a_n=-\frac{1}{n \pi} \int_{-L}^L f'(x) \sin{\frac{n \pi x}{L}} dx$ and $b_n=-\frac{1}{n \pi } \left[ f(L) \cos{n \pi}-f(-L) \cos{n \pi}-\int_{-L}^L f'(x) \cos{\frac{n \pi x}{L}} dx\right]$.

So we get that


$na_n=-\frac{1}{\pi} \int_{-L}^L f'(x) \sin{\frac{n \pi x}{L}} dx$ and $nb_n=-\frac{1}{\pi} \left( f(L) \cos{n \pi}-f(-L) \cos{n \pi}-\int_{-L}^L f'(x) \cos{\frac{n \pi x}{L}} dx\right)$