Stuck in the proof that if $f\in L^1(\mathbb{T})$ and $\hat{f}\in\ell^1(\mathbb{Z})$, then $f\in L^2(\mathbb{T})$

Question

As the title states, I'm stuck showing that, if $f\in L^1(\mathbb{T})$ has absolutely summable Fourier coefficients, then $f$ is square integrable in the torus (wrt Lebesgue measure). The book I'm reading (Folland's Real Analysis) states:

Since the family of exponentials $\{E_n\}_{n\in\mathbb{Z}}$ with $E_n(x):=e^{inx}$ form an orthonormal basis for $L^2(\mathbb{T})$, and since $\ell^1\subset\ell^2$, it follows that $f\in L^2(\mathbb{T})$.

I can't see why this is the case. I know that the map $h\mapsto \hat{h}$ is an isometry from $L^2(\mathbb{T})$ onto $\ell^2(\mathbb{Z})$, and hence $\|h\|_2=\|\hat{h}\|_2$. If we could apply this last equality (Parseval's identity), then is is immediate that $f\in L^2$, but to use it, we need the very thing we are trying to prove, which seems circular. How do we go around this?

EDIT: To clarify, I can see that since $\hat{f}\in\ell^2$, the function $g$ defined by $g(x):=\sum \hat{f}(n)e^{inx}$ belongs to $L^2$ and has the same Fourier series as $f$. We would be done if we could should $f=g$ a.e. but I can't see how to do this. Since $\mathbb{T}$ has finite measure, $L^2\subset L^1$, and hence $g\in L^1$, so it suffices to show $\|f-g\|_1=0$.

Answer 1

Note that if $g \in L^1(\mathbb{T})$ and $\int g e_n = 0$ for all $n$ (with $e_n(t) = e^{i n t}$) then $g = 0$ ae. (This is straightforward to prove using the fact that the continuous functions are dense in $L^1(\mathbb{T})$.)

Since $n \mapsto \hat{f_n} \in l_1 \subset l_2$, then $\phi = \sum_n \hat{f_n} e_n \in L^2(\mathbb{T}) \subset L^1(\mathbb{T})$.

All we need to do is to show that $f=\phi$ ae.

Since $\int (f-\phi) e_n = 0$ for all $n$, we have $f-\phi = 0$ ae. as desired.

Answer 2

You really just need that if $\hat f ∈ \ell ^2$ then the function defined by $f:=\sum_{k∈\mathbb Z}\hat f_k E_k$ is a well defined $L^2$ function on $\Bbb T$ with Fourier transform $\hat f$. That is, the isometry between $\ell^2$ and $L^2$ is a two-way street: given an $L^2$ function, we have an $\ell^2$ sequence, and vice-versa.

In some detail, for any sequence with the suggestive name $\hat f ∈ \ell^2$, we can define a function $f:=\sum_{k∈ \Bbb Z} \hat f_k E_k ∈ L^2(\Bbb T)$, where $E_k$ are the Fourier basis. By orthogonality of $E_k$, $‖f‖^2_{L^2(\Bbb T)} = \sum |\hat f_k|^2 = ‖\hat f‖^2_{\ell^2} $. Also, if $\mathcal F:L^2 → \ell^2$ is the Fourier Transform then an easy check that for this particular $f$ defined above, we can recover $\hat f= \mathcal F f $.

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Response to comments in reverse order -

The series converges in the $L^2$ sense; some work is hidden behind the words 'well-defined'. One needs to first check that $ \{ ∑_{|k|<N} \hat f_k E_k \}_{N\geq 0}$ is Cauchy in the $L^2$ norm; as $L^2$ is complete, there is a limit and we are free to define $$f:=\lim_{N→∞,\ L^2} \sum_{|k|<N} \hat f_k E_k$$

The $f$ I defined coincides with the sum purely by its definition; I suppose you mean to ask if you start with $f ∈ L^2$ and create $\mathcal F f ∈ \ell^2$, then how do we know that $f = ∑_{k∈ \Bbb Z} (\mathcal F f)_k E_k$? This is exactly because $E_k$ form an orthogonal basis, so that $⟨ f, E_k ⟩ =0 $ for every $k$ implies $f=0$.

A proof of this fact is based on the Weierstrass approximation theorem, which says that continuous functions $C$ on the circle can be uniformly approximated by (finite) trigonometric polynomials $\{ ∑_{|k|≤n}a_k E_k : a_k ∈ \Bbb R,\ n = 0,1,2,…\}$. [the version on a closed interval implies the version on the circle by mapping $[0,2\pi]→ \Bbb T$ with $E_1(x) = e^{ix}$.]

At the same time, $L^2$ is the completion of $C$ under the $L^2$ norm. So if we approximate $f ∈ L^2$ by a continuous periodic $g ∈ C(\Bbb T)$, and approximate $g$ uniformly by a finite trigonometric polynomial $g_N$, we obtain with careful $ε$ management,

$$ ‖f - g_N ‖_{L^2} ≤ ‖f - g‖_{L^2} + ‖g - g_N‖_{L^2} < ε $$

By the hypothesis $0 = ∑_{k=-N}^N ⟨f,E_k⟩ E_k$. But also, this sum defines the orthogonal projection i.e. closest point (in $L^2$ distance) to $f$ in $\operatorname{span} \{E_{-N},…,E_{N}\}$. Hence $$‖f-0‖_{L^2} ≤ ‖f - g_N ‖_{L^2} ≤ ε $$ Since $ε>0$ was arbitrary, $f = 0$.

If we now apply this result to $f -\sum_{k∈\Bbb Z} (\mathcal F f)_k E_k$, the result follows.