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For the functions on $[0, 1]$ we have that $L ^q \subset L^p$ for $1 \leq p < q \leq \infty$. Also consider $f_n = x^n$. This sequence converges to 0 in the $L^1$ norm but not $L^\infty$ norm. These statements seem in conflict to me: if $L^1([0,1]) \supset L^\infty ([0,1])$ shouldn't convergence in $L^1$ imply convergence $L^\infty$?

yoshi
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No.

The inclusion you wrote means that there are elements of $L^1([0,1])$ which are not contained in $L^{\infty}([0,1])$, but it doesn't tell us about the topology of the spaces. One such element is $f\colon x\mapsto x^{-1/2}$. Thus $L^{\infty}([0,1])$ is smaller than $L^1([0,1])$ based on this inclusion.

In terms of convergence, the topology on $L^{\infty}([0,1])$ is stronger (or finer) than the topology on $L^{1}([0,1])$. If the sequence $f_n$ converges in $L^{\infty}([0,1])$ then, for all $\epsilon>0$, there is $N$ such that $n>N$ implies

$$\sup_{x\in[0,1]}|f_n(x)-f(x)|<\epsilon.$$

This of course implies that $\int_0^1 |f_n(x)-f(x)|dx <\epsilon$ since $\int_0^1|g(x)|dx \leq \sup_{x\in[0,1]}|g(x)| \cdot (1-0)$. The converse, however, is not true.

Jürgen Sukumaran
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    I guess, when I see subset -- I think of the subset inheriting the topology from the larger set. So I guess the subspace topology on the $L^\infty$ subset is different from the usual definition of the $L^\infty$ norm? – yoshi Jun 12 '21 at 14:40
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    As far as I can tell, OP did not point out any element in $L^1([0,1])$ which does not lie in $L^\infty([0,1])$. – Torsten Schoeneberg Jun 12 '21 at 15:16
  • Wait: I'm reading that Fourier coefficients converge in $L^2$ but not $L^1$ (using compact interval $[0, 2\pi]$. If convergence in $L^2$ implies convergence in $L^1$ shouldn't Fourier coefficients converge in $L^1$? (see link) https://en.wikipedia.org/wiki/Convergence_of_Fourier_series#Norm_convergence – yoshi Jun 12 '21 at 16:13
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    If the partial sums (not the coefficients btw) converge to $f$ in $L^2([0,1])$ and $f$ is $L^2([0,1])$ integrable then the partial sums converge to $f$ in $L^1([0,1])$. The statement they're making is different, they are considering $f$ to only be in $L^1([0,1])$ now which changes things because there are functions in $L^1([0,1])$ which are not in $L^2([0,1])$. – Jürgen Sukumaran Jun 12 '21 at 17:16
  • Can you provide more detail (or reference) about the first statement? If $f \in L^2$ and Fourier partial sums converge $f$, then Fourier partial sums converge to $f$ in $ L^1$ sense? – yoshi Jun 12 '21 at 19:18
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    https://math.stackexchange.com/questions/1529156/uniform-convergence-implies-l2-convergence-and-l2-convergence-implies-l1/1529187

    More generally, for $1\leq p < q \leq \infty$, $L^q$ convergence implies $L^p$ convergence (when you are on a compact set). It's not specific to Fourier series, if you have any sequence of functions converging in a strong topology it will also converge in a weaker topology.

     – Jürgen Sukumaran Jun 12 '21 at 21:46