Uniform Convergence

Question

Let $f_n= \frac{n+cosx}{2n+sin^2x}$.

(a)Prove $f_n \to \frac{1}{2}$ uniformly on $\mathbb{R}$.

(b)Calculate $\displaystyle{\lim_{n\to\infty}} \int^1_7 f_n(x)dx$.

For (a) I am not sure how to proceed. I can't assume it is pointwise convergent so i cant say $f(x) = 1/2$, can I? I can't see how the M-test would help.

I was thinking of using contrapositive the corollary:

If {$f_n$} is a sequence of continuous real value functions on $E$, and if {$f_n$} converges uniformly to $f$ on E, then $f$ is continuous on $E$.

But that won't help me prove convergence will it? Is the converse of this true?

For (b) We can take the limit inside the integral (once we prove (a)) so that just goes to $3$ via a theorem that we prove (no name was given). Right?

Answer 1

Explain why over all $\Bbb R$, $$\left| {{f_n} - \frac{1}{2}} \right| = \left| {\frac{{n + \cos x}}{{2n + {{\sin }^2}x}} - \frac{1}{2}} \right| = \frac{1}{2}\left| {\frac{{2\cos x - {{\sin }^2}x}}{{2n + {{\sin }^2}x}}} \right| \leqslant \frac{3}{2}\frac{1}{{2n}}$$

This gives that $$\lVert f_n-1/2\rVert_\infty \leqslant \frac{3}{4n}\to 0$$

Answer 2

You don't need to assume pointwise convergence. If the exercise only said "prove that $f_n$ converges uniformly on $\mathbb R$", then you'd probably need to prove (or at least convince yourself) that it converges pointwise to $\frac{1}{2}$, or else the proof for uniform convergence would be harder. To solve the exercise, the fact that $|\sin x| \leq 1$ and $|\cos x| \leq 1$ might be helpful.

The corollary you cited is helpful for proving that a succession does not converge uniformly.

As for (b), your idea is correct.