Let R be the set of real numbers. Suppose f : R → R be a continuous and periodic function with period T > 0. Prove that for every a < b,
\lim_{n\to \infty} \int_a^b f(nx)\ \mathrm dx = \dfrac{b-a}{T}\int_0^T f(x)\ \mathrm dx
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2 Answers
1I am trying to give a rather tentative solution .
" 1T 0∫T f ( x ) d x " is the average value of the function " f ( x ) " .
Now , a∫b f ( n x ) dx
= 1n a n∫b n f ( z ) dz
The Mean value of " f ( z ) " is given by ,
M . V = ( b n - a n ) f ( c )n , where " c " is a number between " a n " and " b n " .
= ( b - a ) f ( c )
Now , as n approaches infinity , " c " approaches infinity as well , but as " f ( x ) " is a periodic
function, so if " c " approaches infinity , then " f ( c ) " oscillates infinitely many times . So to find
" f ( c ) " , we can employ the tool " averaging " , i . e , we find out the average of the function in
the extreme case .
So , " f ( c ) " can be viewed as the average of the function " f ( x ) " .
Hence , Given limit = ( b - a ) 1T 0∫T f ( x ) d x
