# Uniform Convergence

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?

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

• How did you get the $\frac{3}{2}$? Feb 23, 2014 at 1:41
• and the \frac{1}{2n}? Wouldn't it be $\frac{1}{2}\frac{1}{2n+1}$? Feb 23, 2014 at 1:49
• @Skuttle_Butt $|2 \cos x - \sin^2 x| \leq |2 \cos x| + |\sin^2 x| \leq 2 + 1 = 3$. Also $2n + \sin^2 x \geq 2n$, therefore $\displaystyle \frac{1}{2n + \sin^2 x} \leq \frac{1}{2n}$. Feb 23, 2014 at 2:20
• Gotcha, thanks. This doesn't show that it converges to $\frac{1}{2}$ though does it? We could have easily used $\frac{1}{3}$ and gotten the same result couldn't we? Feb 23, 2014 at 2:34
• @Skuttle_Butt No, because the $n$ in the denominator would not have cancelled. Limits are unique.
– Pedro
Feb 23, 2014 at 2:35

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.