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NOTE- The source of question is Advanced Calculus on real axis which doesn't cover $\textsf{Lebesgue integral}$ and $\textsf{Measure Theory}$ so therefore an approach by those wouldn't be of much help to me (atleast till the time I haven't gone through those)

This post has been suggested as a duplicate but the two answers posted rely on $\textsf{Measure Theory}$

Let$\ \ f:[0,1]\to \mathbb{R}$ be a continuous function. Show that $$\lim_{\lambda\to\infty}\int_0^1 f(x)\sin(\lambda x)\,dx = 0.$$

Here $f$ is only given to be continuous if $f$ was given to be differentiable then we might have simply used integration by part and bounded it

$ \left| {\int_0^1 f(x)\sin(\lambda x)\,dx}\right | = \left| {\left[ -\frac{f(x)}{\lambda}\cos(\lambda x) \right]_0^1 + \frac{1}{\lambda}\int_0^1 f'(x)\cos(\lambda x)\,dx} \right| \le \frac M{\lambda}$

now $\left[ -\frac{f(x)}{\lambda}\cos(\lambda x) \right]_0^1$ is finite. for the second part as $\cos \lambda x \le1$ and $f'(x)$ is finite because $f(x)$ is defined on a closed and bounded interval therefore it is only finite, if we take the limit to infinity the RHS is $0$.

with the differentiabilty condition I've not been able to think of any strategy to solve it. There is a similar post like this but the problem and the solution are well beyond my current scope.

Edit- based on the hint given by Kavi Ram Murthy I have posted a solution but I don't know if it's entirely correct or not.

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    $\begingroup$ This has been asked a lot of times $\endgroup$ Commented Dec 4 at 7:55
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    $\begingroup$ Approximate $f$ by a function which takes constant values in the intervals $(\frac {i-1}n,\frac i n]$ for each $i$. For such functions the result is easy to verify by direct computation. $\endgroup$ Commented Dec 4 at 7:56
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    $\begingroup$ @JCQ I believe these questions similar in the sense that technically speaking, if the statement in the linked question is proven, then the statement in this here question is also proven because continuous functions on [0,1] are also L1. But they are very different in the sense that an answer to this question here takes much less advanced methods to answer (you can use the uniform continuity of continuous functions on compact intervals), which yields interesting answers that couldn't be found at the linked question. $\endgroup$ Commented Dec 4 at 8:58
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    $\begingroup$ If you want to directly apply famous theorem, you can either using Riemann-Lebesgue Lemma as any continuous function are in $L^1[0,1]$ second way is to use Bessel inequality as any continuous function is in $L^2[0,1]$. $\endgroup$ Commented Dec 4 at 9:05
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    $\begingroup$ Here's an idea. Why not close that PSQ question instead of this one? $\endgroup$ Commented Dec 4 at 9:59

1 Answer 1

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Based on the hint given by Kavi Ram Murthy

Fix $\varepsilon>0$. Since $f$ is uniformly continuous on $[0,1]$, there exists $n\in\mathbb{N}$ such that $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|\le \frac{1}{n}$.

For this $n$, define a step function $f_n:[0,1]\to\mathbb{R}$ by $f_n(x)=f\!\left(\frac{i}{n}\right)$ for $x\in\left(\frac{i-1}{n},\frac{i} {n}\right]$, $i=1,\dots,n$, and set $f_n(0)=f(0)$.

If $x\in\left(\frac{i-1}{n},\frac{i}{n}\right]$, then $\left|x-\frac{i} {n}\right|\le \frac{1}{n}$, hence by the choice of $n$ we have

$|f(x)-f_n(x)| = \left|f(x)-f\!\left(\frac{i}{n}\right)\right| < \varepsilon$.

Therefore $\max_{x\in[0,1]} |f(x)-f_n(x)| \le \varepsilon$.

Define

$I(\lambda)=\int_0^1 f(x)\sin(\lambda x)\,dx$ and

$I_n(\lambda)=\int_0^1 f_n(x)\sin(\lambda x)\,dx$.

Then $I(\lambda)-I_n(\lambda) =\int_0^1 (f(x)-f_n(x))\sin(\lambda x)\,dx$,

so $|I(\lambda)-I_n(\lambda)| \le\int_0^1 |f(x)-f_n(x)|\,|\sin(\lambda x)|\,dx \le\int_0^1 \varepsilon\cdot 1\,dx =\varepsilon.$ $\tag 1$

Nw since $f_n$ is constant on each subinterval,

$I_n(\lambda) =\sum_{i=1}^n f\!\left(\frac{i}{n}\right)\int_{\frac{i-1}{n}}^{\frac{i}{n}} \sin(\lambda x)\,dx$.

we obtain $I_n(\lambda) =\frac{1}{\lambda}\sum_{i=1}^n f\!\left(\frac{i}{n}\right) \left(\cos\!\left(\lambda\frac{i-1}{n}\right)-\cos\!\left(\lambda\frac{i}{n}\right)\right)$.

Let $M=\max_{x\in[0,1]} |f(x)|$. Since $|\cos(\cdot)|\le 1$, we have $\left|\cos\!\left(\lambda\frac{i-1}{n}\right)-\cos\!\left(\lambda\frac{i}{n}\right)\right|\le 2$, hence

$|I_n(\lambda)| \le\frac{1}{|\lambda|} \sum_{i=1}^n \left|f\!\left(\frac{i}{n}\right)\right|\cdot 2 \le\frac{1}{|\lambda|}\cdot 2nM =\frac{2nM}{|\lambda|}$.

For this fixed $n$ we therefore have $\lim_{\lambda\to\infty} I_n(\lambda) = 0$. Thus there exists $\Lambda>0$ such that for all $|\lambda|\ge\Lambda$, $|I_n(\lambda)|\le\varepsilon.$ $\tag 2$

Combining $(1)$ and $(2)$, for all $|\lambda|\ge\Lambda$ we get

$|I(\lambda)| \le |I(\lambda)-I_n(\lambda)| + |I_n(\lambda)| \le \varepsilon + \varepsilon = 2\varepsilon$.

Since $\varepsilon>0$ was arbitrary, it follows that $\lim_{\lambda\to\infty}\int_0^1 f(x)\sin(\lambda x)\,dx = 0$.

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  • $\begingroup$ Looks fine. I think the second part could be a separate question and make it explicit that Measure Theory should not be used. $\endgroup$ Commented Dec 4 at 9:50
  • $\begingroup$ @KaviRamaMurthy thanks this idea of usig step function was new to me $\endgroup$ Commented Dec 4 at 9:51

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