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Aug 24, 2010 at 23:20 comment added Dorian I'm concerned with $f \in L^1([0,1])$ but not in $L^2([0,1])$ so that you can't just use the $L^2$ structure.
Aug 24, 2010 at 19:38 comment added Otis Chodosh By Holder's inequality on $[0,1]$, $\Vert f \Vert_{L^1} \leq \Vert f \Vert_{L^2}$ so what I have shown is that if you started with a f that was actually in L2 , then you can replace the L2 convergence statements with L1 convergence.
Aug 24, 2010 at 18:59 comment added Dorian You hae misunderstood my question. I'm concerned with $L^1([0,1])$. Of course we have a fourier series in $L^2([0,1])$ but my point is that we get coefficients regardless and we should get convergence in $L^1([0,1])$ as well for certain coefficients (not the fourier coefficients though a priori if our function is not $L^2([0,1])$).
Aug 24, 2010 at 18:43 history answered Otis Chodosh CC BY-SA 2.5