continuity of sequence of functions

Question

For each $n\in\mathbb{N}$, define $f_{n}:\mathbb{R}\to\mathbb{R}$ by $$f_{n}(x)=\int_{0}^{n}\frac{\sin(xt)}{e^{t}+x^{2}}\,dt.$$

How to ensure that each $f_{n}$ is continuous on $\mathbb{R}$ by definition of the continuity?

I've tried as follows:

To letting $h\to0$ at last, for any $h\neq0$, small enough, $$|f_{n}(x+h)-f(x)|\le\int_{0}^{n}\left|\frac{\sin((x+h)t)}{e^{t}+(x+h)^{2}}-\frac{\sin(xt)}{e^{t}+x^{2}}\right|dt.$$

In the latter expression, the magnitude of integrand depends on the $h$'s sign.

So, i can't go any further. Give some advice. Thank you!

Answer 1

Hints: $\frac {\sin ((x+h)t)} {e^{t}+(x+h)^{2}} -\frac {\sin (xt)} {e^{t}+x^{2}}$ can be split into wto terms: $\frac {\sin ((x+h)t)} {e^{t}+(x+h)^{2}} -\frac {\sin ((x+h)t)} {e^{t}+x^{2}}$ and $\frac {\sin ((x+h)t)} {e^{t}+x^{2}} -\frac {\sin (xt)} {e^{t}+x^{2}}$. In the first of these terms use the fact that $|sin ((x+h)t)| \leq 1$ and $|\frac 1 {e^{t}+(x+h)^{2}} -\frac 1 {e^{t}+x^{2}}| \leq | {h^{2}+2hx}|$ since $e^{t}+x^{2} \geq 1$ and $e^{t}+(x+h)^{2} \geq 1$. In the second term you can again use the fact that the denominator is $ \geq 1$. So you are left with $|sin((x+h)t-\sin(xt)|$. You can use the fact that $\sin((x+h)t)=\sin xt \cos (ht)+\cos xt \sin (ht)$ to handle this. I hope you can handle the rest of the argument.

Answer 2

If you know the dominated convergence theorem (DCT) for integrals you get the continuity of the function $f_n$ by applying the characterization of continuity by sequences. To show is the following

• For any $x_0 \in \mathbb{R}$ and any sequence $(x_k)_{k\in\mathbb{N}}$ with $x_k \in \mathbb{R}$ and $\lim_{k\to \infty}x_k = x_0$ you have $$\lim_{k \to \infty}f_n(x_k) = f_n(x_0)$$

But this follows directly from above mentioned theorem as follows:

• Set $g_k(t) =\frac{\sin(x_kt)}{e^{t}+x_k^{2}}$, then because of the continuity of the given function you have $$g_k(t) \stackrel{k\to \infty}{\longrightarrow}g_0(t) = \frac{\sin(x_0t)}{e^{t}+x_0^{2}}$$
• You have $|g_k(t)| \leq \frac{1}{e^t}$, which is an integrable majorant on any interval $[0,n]$. So, you get $$f_n(x_k) = \int_{0}^{n}g_k(t)dt \stackrel{k\to \infty(DCT)} {\longrightarrow} \int_{0}^{n}g_0(t)dt = f_n(x_0)$$

This is the required result.

Answer 3

Fix $x\in \mathbb{R}$. The function $h: [0,n]\times [-2\delta,2\delta]\to \mathbb{R}$ defined by $g(h,t)=\frac{\sin((x+h)t)}{e^{t}+(h+x)^{2}}$ is uniformly continuous, as a continuous function defined on a compact set.

Therefore there is a $\delta $ such that $ |g(v_1)-g(v_2)|<\epsilon/n $ for all $v_1,v_2$ such that $||v_1-v_2||_2 < \delta$.

So, $ |g(x+h,t)- g(x,t )|<\epsilon/n$ for all $(t,h)\in [0,n]\times (\delta,\delta)$ since $||(x+h,t)- (x,t )||_2=|h|<\delta$

Hence,

\begin{align} |f_n (x+h)-f_n(x)| &\le\int_{0}^{n}\left|\frac{\sin((x+h)t)}{e^{t}+(x+h)^{2}}-\frac{\sin(xt)}{e^{t}+x^{2}}\right|dt\\ &=\int_{0}^{n}\left|g(x+h,t)- g(x,t )\right|dt\\ &<\int_{0}^{n}\epsilon/ndt\\ &=\epsilon \end{align}