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Let $f_n:[0,1]\to\mathbb{R}$ be a sequence continuous functions which uniformly converges to continuous function $f:[0,1]\to\mathbb{R}$.

Let

$$F_n=\int_0^1\frac{\sin(nx)}{2+\cos(nx)}f_n(x)dx,\:\:\:n\geq 1.$$

Is $F_n$ necessarily convergent?

As $f_n\to f$ uniformly, the sequence is uniformly bounded so $\frac{\sin(nx)}{2+\cos(nx)}f_n(x)$ is bounded. I'm stuck on finding the limit of $\frac{\sin(nx)}{2+\cos(nx)}f_n(x)$, if it has a Riemann integrable limit, then by Dominated convergent theorem, I can pass the limit inside the integral, I guess! However, I'm not sure about using the dominated convergent theorem I appreciate any help.

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Given a partition $0 = x_0 < x_1 < \ldots < x_m = 1$, we have

$$\left|\int_0^1 \frac{\sin nx}{2 + \cos nx } f_n(x) \, dx \right| \\ \leqslant \sum_{j=1}^m \int_{x_{j-1}}^{x_j} |f_n(x) - f_n(x_j)|\frac{|\sin n x|}{|2 + \cos nx|} \, dx + \sum_{j=1}^m |f_n(x_j)| \left|\int_{x_{j-1}}^{x_j} \frac{\sin n x}{2 + \cos nx} \, dx\right| \\ \leqslant \sum_{j=1}^m \int_{x_{j-1}}^{x_j} |f_n(x) - f_n(x_j)| \, dx + \sum_{j=1}^m |f_n(x_j)| \frac{|\log(2+\cos nx_j)- \log(2+ \cos n x_{j-1})|}{n} \\ \leqslant \sum_{j=1}^m \int_{x_{j-1}}^{x_j} |f_n(x) - f_n(x_j)| \, dx + \sum_{j=1}^m |f_n(x_j)| \frac{\log(3)}{n}$$

The sequence $f_n$ is uniformly bounded and there exists $M > 0$ such that $|f_n(x)| \leqslant M$ for all $x \in [0,1]$ and for all $n$.

We can use uniform convergence of $f_n$ and uniform continuity of the limit function $f$ to find $n$ and $m$ sufficiently large (and sufficiently fine partition norm ) such that $|f_n(x) - f_n(x_j)| < \epsilon/2$ for all $x \in [0,1]$. The argument uses the estimate

$$|f_n(x) - f_n(x_j)| \leqslant |f_n(x) - f(x)| + |f(x) - f(x_j)| + |f(x_j) - f_n(x_j)|.$$

Hence,

$$\left|\int_0^1 \frac{\sin nx}{2 + \cos nx } f_n(x) \, dx \right| \leqslant \frac{\epsilon}{2} \sum_{j=1}^m\int_{x_{j-1}}^{x_j} dx + \frac{\log(3)Mm}{n} = \frac{\epsilon}{2} + \frac{\log(3)Mm}{n}.$$

For all sufficiently large $n$, the RHS is less than $\epsilon$ and it follows that

$$\lim_{n \to \infty}\int_0^1 \frac{\sin nx}{2 + \cos nx } f_n(x) \, dx = 0$$

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We will prove a more general fact, from which you should be able to deduce the limit on your own.

Claim: Suppose that $g$ is periodic and continuous with period $\alpha>0$. Then for any sequence of continuous functions $f_n \to f$ uniformly, we have that $$\int_0^1g(nx)f_n(x)dx \stackrel{n \to \infty}{\longrightarrow} \bigg(\frac{1}{\alpha}\int_0^{\alpha} g(x)dx \bigg) \cdot \int_0^1f(x)dx $$

Proof: This fact can basically be seen as some version of the ergodic theorem, but in this case we can give an elementary proof: Define $\tilde f_n(x)$ to be $f(k \alpha / n)$ whenever $x \in [k\alpha/n, (k+1)\alpha/n) \cap [0,1]$, with $k \in \Bbb Z_{\geq 0}$. So each $\tilde f_n$ is a piecewise constant function.

Then I claim that $\tilde f_n \to f$ uniformly. Indeed, for $\epsilon>0$ take $N$ large enough so that $|f(x)-f(y)| \leq \epsilon$ when $|x-y| \leq \alpha N^{-1}$ (possible by uniform continuity). Then for $n > N$ we have that for every $x$, say $x \in [k\alpha/n, (k+1)\alpha/n)$, that $|f(x) - \tilde f_n(x)| =|f(x)-f(k\alpha/n)|< \epsilon$. This proves that $\tilde f_n \to f$ uniformly, which easily yields (since $g$ is bounded) that $$\int_0^1 |g(nx)| |\tilde f_n(x)-f_n(x)| dx \leq \int_0^1 |g(nx)| |\tilde f_n(x)-f(x)|dx+\int_0^1 |g(nx)| | f_n(x)-f(x)|dx \stackrel{n \to \infty}{\longrightarrow} 0$$ where we used the fact that $\tilde f_n\to f$ uniformly and $f_n \to f$ uniformly.

Now all we have to do is find $\lim_{n \to \infty} \int_0^1 g(nx)\tilde f_n(x)dx$. But that's not difficult: \begin{align*} \int_0^1 g(nx) \tilde f_n(x) &= \sum_{k=0}^{\lfloor n /\alpha \rfloor} \int_{k\alpha/n}^{(k+1)\alpha/n} g(nx)f(k\alpha/n) dx + \int_{\lfloor n /\alpha \rfloor \alpha/n}^1 g(nx)\tilde f_n(x)dx \\ &= \bigg( \sum_{k=0}^{\lfloor n /\alpha \rfloor} f(k\alpha/n) \bigg) \bigg( \frac{1}{n} \int_0^{\alpha}g(x)dx\bigg) + O(1/n) \\ &= \bigg( \frac{1}{n/\alpha}\sum_{k=0}^{\lfloor n /\alpha \rfloor} f(k\alpha/n) \bigg) \bigg( \frac{1}{\alpha}\int_0^{\alpha}g(x)dx\bigg) + O(1/n) \\ &\stackrel{n \to \infty}{\longrightarrow}\bigg(\int_0^1f(x)dx \bigg) \bigg( \frac{1}{\alpha}\int_0^{\alpha} g(x)dx \bigg) \end{align*} where we recognized a Riemann sum approximation in the final limit. This proves the lemma.

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  • $\begingroup$ Additional hint: by substituting $u=\cos x$, one may see that $\frac{1}{2\pi} \int_0^{2\pi} \frac{\sin x}{2+\cos x} dx =0$ $\endgroup$ – Shalop Sep 4 '17 at 6:44

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