$f_n$ uniformly converge to $f$ and $g_n$ uniformly converge to $g$ then $f_n \cdot g_n$ uniformly converge to $f\cdot g$

Question

Please help me check, if

1. $f_n$ uniformly converge to $f$ and $g_n$ uniformly converge to $g$ then $f_n + g_n$ uniformly converge to $f+g$

2. $f_n$ uniformly converge to $f$ and $g_n$ uniformly converge to $g$ then $f_n \cdot g_n$ uniformly converge to $f\cdot g$

What I've done:

I Guess that first statement is true, second is false. I've tried to calculate directly from definition, nothing seems correct.

Answer 1

For the second case, take $f_n(x)=g_n(x) = x+\frac{1}{n}$. With $f(x)=g(x) = x$, we see that $f_n \to f, g_n \to g$ uniformly.

However, $f_n(x) g_n(x) = f(x)g(x) + \frac{2}{n} x + \frac{1}{n^2}$, hence the convergence is not uniform on unbounded sets.

Answer 2

The second is not corrent try $f_n(x)= \frac{1}{n}$ and $g_n(x)=x$ which both converge uniformly. Now their product is $h_n(x)= \frac{x}{n}$ which converge pointwise to the zero function. But $$\sup _{x \in \mathbb{R}}|h_n(x)-0|=\infty \forall n\in \mathbb{N}$$ Therefore it cannot converge uniformly to the zero function.

Notice that the only thing that was keeping you from applying the triangle inequality was the boundness of $f_n$ and $g_n$.

Answer 3

Let $\epsilon >0$ there's $n_1\in \mathbb{N}$, $\forall n\geq n_1$ $||f_n-f||_\infty<\epsilon/2$ and there's $n_2\in \mathbb{N}$, $\forall n\geq n_2$ $||g_n-g||_\infty<\epsilon/2$,

now let $n_0=\max(n_1,n_2)$ then $\forall n\geq n_0$ $||f_n+g_n-(f+g)||_\infty\leq ||f_n-f||_\infty+||g_n-g||_\infty <\epsilon$

For the second result if we suppose that the functions are continuous on the compact $[a,b]$ then use the same idea with the fact that $(f_n)$ is bounded since it's convergent sequence and with the inequality $$||f_n g_n - fg||_\infty=||f_n g_n - f_n g + f_n g -fg||_\infty\leq ||f_n||_\infty||g_n-g||_\infty + ||g||_\infty||f_n-f||_\infty$$