A Fourier Analysis Question I am stuck at

Question

If $f,g\in C[-\pi,\pi]$,and $f,g$ are $2\pi$ periodic, prove that $$\lim_{n\to\infty}\dfrac{1}{2\pi}\int_{-\pi}^\pi f(t)g(nt)\mathbb dt=\big(\dfrac{1}{2\pi}\int_{-\pi}^\pi f(t)\mathbb dt\big)\big(\dfrac{1}{2\pi}\int_{-\pi}^\pi g(t)\mathbb dt\big)$$

I tried to approximate $g(nt)$ by a trigonometric polynomial $T_n(t)$ but the problem is that $n$ is varying here. Nothing seems to work. I tried to interpret the left side as the constant Fourier coefficient of the function $f(x)g(nx)$ and the right side as the product of constant Fourier coefficients of $f(x)$ and $g(x)$ but because $f$ and $g$ are just continuous and not Lipschitz, I cannot replace $f$ and $g$ by their Fourier sums.

Please give me a hint only to get started on this problem.

Answer 1

Let $a_{k}$ and $b_{k}$ be $$ a_{k}=\frac{1}{2\pi}\int_{0}^{2\pi}e^{-ikx}f(x)dx,\;\;\; b_{k}=\frac{1}{2\pi}\int_{0}^{2\pi}e^{-ikx}g(x)dx. $$ Then $g(x)=\sum_{k=-\infty}^{\infty}b_{k}e^{ikx}$ converges in $L^{2}$ on any finite interval to the periodic extension of $g$. More explicitly, if $S_{N}=\sum_{n=-N}^{N}b_{k}e^{ikx}$, then, for fixed $n=1,2,3,\cdots$, $$ \lim_{N\rightarrow\infty}\int_{0}^{2n\pi}|g(x)-S_{N}(x)|^{2}dx= 0. $$ By a simple change of variables, for fixed $n=1,2,3,\cdots$, one has $$ \lim_{N\rightarrow\infty}\int_{0}^{2\pi}|g(nx)-S_{N}(nx)|^{2}dx = 0. $$ Then, for $f \in L^{2}$, it follows that, for fixed $n=1,2,3,\cdots$, $$ \int_{0}^{2\pi}f(x)g(nx)dx = \lim_{N\rightarrow\infty}\int_{0}^{2\pi}f(x)S_{N}(nx)dx. $$ The above is easily justified using Cauchy-Schwarz: $$ \left|\int_{0}^{2\pi}f(x)g(nx)dx-\int_{0}^{2\pi}g(x)S_{N}(nx)dx\right| \\ \le \int_{0}^{2\pi}|f(x)||g(nx)-S_{N}(nx)|dx \\ \le \left[\int_{0}^{2\pi}|f(x)|^{2}\right]^{1/2} \left[\int_{0}^{2\pi}|g(nx)-S_{N}(nx)|^{2}\right]^{1/2}\rightarrow 0 \mbox{ as } N \rightarrow \infty \mbox{ for fixed n. } $$

Therefore, for any fixed $n=1,2,3,\cdots,$ $$ \begin{align} \frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(nx)dx & =\lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{0}^{2\pi}f(x)S_{N}(nx)dx \\ & = \lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{0}^{2\pi}f(x)\sum_{k=-N}^{N}b_{k}e^{iknx} dx \\ & = \lim_{N\rightarrow\infty}\sum_{k=-N}^{N}b_{k}\frac{1}{2\pi}\int_{0}^{2\pi}f(x)e^{iknx}dx \\ & = \lim_{N\rightarrow\infty}\sum_{k=-N}^{N}b_{k}a_{-kn} \\ & = \sum_{k=-\infty}^{\infty}b_{k}a_{-kn}. \end{align} $$ Finally, Applying Cauchy-Schwarz to the sum on the right gives $$ \begin{align} \left|\frac{1}{2\pi}\int_{0}^{2\pi}f(x)g(nx)dx-a_{0}b_{0}\right|^{2} & = \left|\sum_{|k| \ge 1}b_{k}a_{kn}\right|^{2} \\ & \le \sum_{|k|\ge 1}|b_{k}|^{2}\sum_{|k|\ge 1}|a_{kn}|^{2} \\ & \le \sum_{|k|\ge 1}|b_{k}|^{2}\sum_{|k| \ge n}|a_{k}|^{2}\rightarrow 0 \mbox{ as } n\rightarrow\infty. \end{align} $$ The last term on the right tends to $0$ because the sum is convergent.

Answer 2

Use the real form of Fourier series. The first thing that we examine is the integral $\frac{1}{{2\pi }}\int_{ - \pi }^\pi {f(t)g(nt)dt} $.

If we can determine this integral, then we can work out the limit. Both f and g are continuous and therefore square integrable. Note that since they are both periodic of period 2$\pi$, the convergence properties of the Fourier series of both functions should be useful. Though their Fourier series converge almost everywhere in $[-\pi, \pi]$, but this property is of little use. But since they are both square integrable, their Fourier series converge respectively to $f$ and $g$ in the L² norm.

Let the Fourier series of $f$ be

$T(\theta ) = \frac{1}{2}{a_0} + \sum\limits_{j = 1}^\infty {\left( {{a_j}\cos (j\theta ) + {b_j}\sin (j\theta )} \right)} $
and the Fourier series of $g$ be

$S(\theta ) = \frac{1}{2}{A_0} + \sum\limits_{j = 1}^\infty {\left( {{A_j}\cos (j\theta ) + {B_j}\sin (j\theta )} \right)} $

and

${T_k}(\theta ) = \frac{1}{2}{a_0} + \sum\limits_{j = 1}^k {\left( {{a_j}\cos (j\theta ) + {b_j}\sin (j\theta )} \right)} $ and

${S_k}(\theta ) = \frac{1}{2}{A_0} + \sum\limits_{j = 1}^{j = k} {\left( {{A_j}\cos (j\theta ) + {B_j}\sin (j\theta )} \right)} $ their respective partial sums.

Then convergence in the L² norm means $\mathop {\lim }\limits_{k \to \infty } \int_{ - \pi }^\pi {{{\left| {{T_k}(t) - f(t)} \right|}^2}} = 0$ and $\mathop {\lim }\limits_{k \to \infty } \int_{ - \pi }^\pi {{{\left| {{S_k}(t) - g(t)} \right|}^2}} = 0$.

Let $n$ be a fixed positive integer.

Then we also have that $\mathop {\lim }\limits_{k \to \infty } \int_{ - n\pi }^{n\pi } {{{\left| {{S_k}(t) - g(t)} \right|}^2}} = 0$.

Thus by a change of variable we get $\mathop {\lim }\limits_{k \to \infty } n\int_{ - \pi }^\pi {{{\left| {{S_k}(nt) - g(nt)} \right|}^2}dt} = 0$ and so
$\mathop {\lim }\limits_{k \to \infty } \int_{ - \pi }^\pi {{{\left| {{S_k}(nt) - g(nt)} \right|}^2}dt} = 0$.

This means ${S_k}(nt)$ tends to $g(nt)$ in the L² norm and we have $\frac{1}{\pi }\int_{ - \pi }^\pi {f(t){S_k}(nt)dt} $ tends to $\frac{1}{\pi }\int_{ - \pi }^\pi {f(t)g(nt)dt}$.

We can deduce this statement as follows: $\left| {\frac{1}{\pi }\int_{ - \pi }^\pi {f(t){S_k}(nt)dt} - \frac{1}{\pi }\int_{ - \pi }^\pi {f(t)g(nt)dt} } \right|$

$ \le \frac{1}{\pi }\int_{ - \pi }^\pi {|f(t)||{S_k}(nt) - g(nt)|dt} $

$ \le \frac{1}{\pi }\sqrt {\left( {\int_{ - \pi }^\pi {|f(t){|^2}dt} } \right)} \sqrt {\int_{ - \pi }^\pi {|{S_k}(nt) - g(nt){|^2}dt} }$ by Holder’s Inequality.

Since $\mathop {\lim }\limits_{k \to \infty } \int_{ - \pi }^\pi {{{\left| {{S_k}(nt) - g(nt)} \right|}^2}dt} = 0$, the right hand side of the above inequality tends to 0 and so by the Comparison Test,

$\frac{1}{\pi }\int_{ - \pi }^\pi {f(t){S_k}(nt)dt} $ tends to $\frac{1}{\pi }\int_{ - \pi }^\pi {f(t)g(nt)dt} $ as $k$ tends to infinity.

We now compute $\frac{1}{\pi }\int_{ - \pi }^\pi {f(t){S_k}(nt)dt} $.

Since $f(t){S_k}(nt) = \frac{1}{2}{A_0}f(t) + \sum\limits_{j = 1}^{j = k} {\left( {{A_j}f(t)\cos (jnt) + {B_j}f(t)\sin (jnt)} \right)} $ ,

$\frac{1}{\pi }\int_{ - \pi }^\pi {f(t){S_k}(nt)dt} = \frac{1}{2}{A_0}\frac{1}{\pi }\int_{ - \pi }^\pi {f(t)dt} + \sum\limits_{j = 1}^{j = k} {\left( {{A_j}\frac{1}{\pi }\int_{ - \pi }^\pi {f(t)\cos (jnt)} dt + {B_j}\frac{1}{\pi }\int_{ - \pi }^\pi {f(t)\sin (jnt)dt} } \right)} $

$ = \frac{1}{2}{A_0}{a_0} + \sum\limits_{j = 1}^{j = k} {\left( {{A_j}{a_{jn}} + {B_j}{b_{jn}}} \right)} $.

It follows that $\frac{1}{\pi }\int_{ - \pi }^\pi {f(t){S_k}(nt)dt} \to \frac{1}{2}{A_0}{a_0} + \sum\limits_{j = 1}^\infty {\left( {{A_j}{a_{jn}} + {B_j}{b_{jn}}} \right)} $ as $k$ tends to infinity.

Hence $\frac{1}{\pi }\int_{ - \pi }^\pi {f(t)g(nt)dt} = \frac{1}{2}{A_0}{a_0} + \sum\limits_{j = 1}^{j = \infty } {\left( {{A_j}{a_{jn}} + {B_j}{b_{jn}}} \right)} $.

Now we claim that $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 1}^\infty {\left( {{A_j}{a_{jn}} + {B_j}{b_{jn}}} \right)} = 0$.

Observe that $\left| {\sum\limits_{j = 1}^{j = k} {\left( {{A_j}{a_{jn}} + {B_j}{b_{jn}}} \right)} } \right| \le \left| {\sum\limits_{j = 1}^{j = k} {{A_j}{a_{jn}}} } \right| + \left| {\sum\limits_{j = 1}^{j = k} {{B_j}{b_{jn}}} } \right| \le \sqrt {\sum\limits_{j = 1}^{j = k} {{A_j}^2} } \sqrt {\sum\limits_{j = 1}^{j = k} {a_{jn}^2} } + \sqrt {\sum\limits_{j = 1}^{j = k} {{B_j}^2} } \sqrt {\sum\limits_{j = 1}^{j = k} {b_{jn}^2} } $

by the Cauchy Schwarz inequality,

$ \le \sqrt {\sum\limits_{j = 1}^\infty {{A_j}^2} } \sqrt {\sum\limits_{j = 1}^\infty {a_{jn}^2} } + \sqrt {\sum\limits_{j = 1}^\infty {{B_j}^2} } \sqrt {\sum\limits_{j = 1}^\infty {b_{jn}^2} } $.

And so $\left| {\sum\limits_{j = 1}^\infty {\left( {{A_j}{a_{jn}} + {B_j}{b_{jn}}} \right)} } \right| \le \sqrt {\sum\limits_{j = 1}^\infty {{A_j}^2} } \sqrt {\sum\limits_{j = 1}^\infty {a_{jn}^2} } + \sqrt {\sum\limits_{j = 1}^\infty {{B_j}^2} } \sqrt {\sum\limits_{j = 1}^\infty {b_{jn}^2} } $

$ \le \sqrt {\sum\limits_{j = 1}^\infty {{A_j}^2} } \sqrt {\sum\limits_{j = n}^\infty {a_j^2} } + \sqrt {\sum\limits_{j = 1}^\infty {{B_j}^2} } \sqrt {\sum\limits_{j = n}^\infty {b_j^2} } $ .

Note that $\frac{1}{\pi }\int_{ - \pi }^\pi {|g(t){|^2}dt} = \frac{1}{2}A_0^2 + \sum\limits_{j = 1}^\infty {\left( {A_j^2 + B_j^2} \right)} $ and $\frac{1}{\pi }\int_{ - \pi }^\pi {|f(t){|^2}dt} = \frac{1}{2}a_0^2 + \sum\limits_{j = 1}^\infty {\left( {a_j^2 + b_j^2} \right)} $.

So both $\sum\limits_{j = 1}^\infty {a_j^2} $ and $\sum\limits_{j = 1}^\infty {b_j^2} $ are finite and as a consequence,

$\sum\limits_{j = n}^\infty {a_j^2} \to 0{\rm{ }}\quad {\rm{and }}\quad \sum\limits_{j = n}^\infty {b_j^2} \to 0{\rm{ }}\quad {\rm{as }}\;n \to \infty $ .

And so it follows from the above inequality that $\mathop {\lim }\limits_{n \to \infty } \sum\limits_{j = 1}^\infty {\left( {{A_j}{a_{jn}} + {B_j}{b_{jn}}} \right)} = 0$.

Hence $\mathop {\lim }\limits_{n \to \infty } \frac{1}{\pi }\int_{ - \pi }^\pi {f(t)g(nt)dt} = \frac{1}{2}{A_0}{a_0} = \frac{1}{{2\pi }}\int_{ - \pi }^\pi {f(t)dt} \frac{1}{\pi }\int_{ - \pi }^\pi {g(t)dt} $,

i.e., $\mathop {\lim }\limits_{n \to \infty } \frac{1}{{2\pi }}\int_{ - \pi }^\pi {f(t)g(nt)dt} = \left( {\frac{1}{{2\pi }}\int_{ - \pi }^\pi {f(t)dt} } \right)\left( {\frac{1}{{2\pi }}\int_{ - \pi }^\pi {g(t)dt} } \right)$.

Observe that I only use the fact that both functions are periodic, Lebesgue integrable and square integrable. The continuity of both $f$ and $g$ is used only to deduce both integrablity and square integrability. We may of course replace the condition on continuity by Lebsgue integrability and square integrability.

Answer 3

Hint: write $g(t)$ as the combination of $\sin(kt)$ and $\cos(kt)$, i.e., you need only solve the problem for $g(t)=\sin(kt)$ or $\cos(kt)$ for some $k$.

Answer 4

I would suggest using that continuous functions on compact sets are uniformly continuous, and then using that $f$ is Riemann integrable to split $f$ up over a partition of $[-\pi,\pi]$.

Alternative suggestion: Fejér sums?

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Okay, I've thought about this some more, and I think it comes down to one of the following lemmata:

Lemma 1: Let $g$ be as in the question, and let $\chi_A$ be the indicator function of a closed subset in $[-\pi,\pi]$. Then $$ \lim_{n \to \infty} \int_{-\pi}^{\pi} \chi_A(t)g(nt) \, dt = \mu(A) \left( \frac{1}{2\pi} \int_{-\pi}^{\pi} g(t) \, dt \right) $$

(From here, I use the usual increasing sequence of step functions argument for Lebesgue integration) (Why do closed sets suffice? For a continuous function, the level sets are closed - see this question.)

or

Lemma 2: Let $g$ be as in the question, and let $\chi_A$ be the indicator function of an interval in $[-\pi,\pi]$. Then $$ \lim_{n \to \infty} \int_{-\pi}^{\pi} \chi_A(t)g(nt) \, dt = \mu(A) \left( \frac{1}{2\pi} \int_{-\pi}^{\pi} g(t) \, dt \right) $$

(This is obviously sufficient to prove the other one, using the Heine-Borel theorem.)

Then you use that and the properties of Riemann sums to sandwich $f$ and take the limit to get the integral. Lemma 2 looks pretty easy to prove.