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Assume now we have $f(x)\in L^1([0,1])$, then we don't necessarily have the convergence of the partial sum of the Fourier series, moreover, by the theorem of kolmogorov, we can even have a.e. divergence of the partial sum.

Now my question is, for $f(x)\in L^1([0,1])$, denote the Fourier Transform as $\{a_n\}_{n=-\infty}^{\infty}$, and assume that the partial sum $S_n(x)=\sum_{k=-n}^{n}a_ke^{ikx}$ converges pointwise almost everywhere in $[0,1]$, then can we expect that the partial sum will converge back to the original function $f(x)$?

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A Fourier transform is different from a Fourier sequence. – Thomas Andrews Oct 17 '11 at 21:57
@Thomas Andrews: but I think in some general sense, there's no problem to call it like this, or? – bonnnnn2010 Oct 17 '11 at 22:00
No, in the "general sense," the Fourier transform of a function is different from the sequence of coefficients of the Fourier series of that function. In mathematics, words are precise things. You can argue, perhaps, that the Fourier series of a function is somehow akin to the Fourier transform, but then the Fourier series is "a (generalized) Fourier transform" of the function, not "the Fourier transform" of that function. – Thomas Andrews Oct 18 '11 at 17:00
up vote 5 down vote accepted

The Cesaro means $\sigma_n(x)$ of the Fourier series of an $L^1$ function $f(x)$ converge almost everywhere to $f(x)$. At any point where the partial sums $S_n(s)$ converge, the Cesaro means $\sigma_n(x)$ converge to the same value. So if the partial sums converge pointwise almost everywhere, the limit must almost everywhere be $f(x)$. Of course, as user16892 noted, the limit might not be $f(x)$ everywhere (but that's obvious, because you can change an $L^1$ function on a set of measure 0 and not affect the Fourier series).

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nice observation, thanks a lot, but pity that I'm not able to vote up – bonnnnn2010 Oct 17 '11 at 22:02

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