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If ${f_k}_{k=1}^\infty$ is a sequence of Reimann Integral functions on [0,1] and $∫_0^1 |f_k(x) - f(x)|dx $ -> 0 as k -> $\infty$. Show $\hat{f}_k(n)$ -> $\hat{f}(n)$ converges uniformly in n as k -> $\infty$

So far a friend of mine told me that $L^1(T)$ convergence of $f_k$ implies uniform convergence of $\hat{f}_k$. But I don't see how that happens.

Any help would be appreciated.

Yeti
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  • have you applied the definition of Fourier coefficients to $|\hat f (n) -\hat f(n)|$. Once you do this and use the inequality $|\int g(x)dx| \leq \int |g(x)|dx$ you will get the result. – Kavi Rama Murthy Mar 22 '18 at 06:09

1 Answers1

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Since the $f_k$ are continuous a.e.

$$|\hat{f}(n)-\hat{f}_k(n)|=\left | \int_0^1 (f(x)-f_k(x))e^{-inx}\right|\le \int_0^1|f(x)-f_k(x)|dx$$

meiji163
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