Convergence of product of sequences of functions

Question

Suppose we have two sequences of functions $(f^1_n),(f^2_n) $ where $f_n^1,f_n^2: \mathbb{R}^n \to \mathbb{R}$.

These sequences verify ($\overset{u}{\rightarrow}$ means uniform convergence): $$f_n^1 g \overset{u}{\rightarrow} g$$

And: $$f_n^2 \overset{u}{\rightarrow} g^2$$ Where $g, g^2:\mathbb{R}^n \to \mathbb{R}$.

Is it true that $f_n^1 f_n^2 \overset{u}{\rightarrow} g_2$?

I am trying to show it by using the definition of uniform convergence but I have not been able to conclude it. That is, if we have $\forall \varepsilon \ \exists n_0$ such that $\forall x \in \mathbb{R}^n$ and all $n \geq n_0$ we have: $$|f_n^1(x)g(x)-g(x)|< \varepsilon $$ $$ |f_n^2(x)-g^2(x)| < \varepsilon $$

Then $|f_n^1(x)f_n^2(x)-g^2(x)|=|f_n^1(x)(f_n^2(x)-g^2(x))+ f_n^1(x)g^2(x)-g^2(x)|\leq |f_n^1(x)(f_n^2(x)-g^2(x))|+ |f_n^1(x)g^2(x)-g^2(x)| < |f_n^1(x)|(f_n^2(x)-g^2(x))| + \varepsilon$

But I am stuck here. If I could show that $f_n$ is bounded I could conclude it, right?

Answer 1

The result is not true in general: take

• $f_n^1 = 0$ for all $n$
• $g=0$
• $g_2$ defined by $g_2(x)=1$ for all $x$
• $f_n^2 = g_2$ for all $n$

Then it is obvious that $$f_n^1 g \overset{u}{\rightarrow} g$$

And: $$f_n^2 \overset{u}{\rightarrow} g_2$$

But $f_n^1 f_n^2= 0$ and $g_2=1$ so $f_n^1 f_n^2$ can't converge uniformly to $g_2$.

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However the result is is true if there exists $\eta >0$ such that $|g(x)| \geq \eta$ for all $x$ and if $g_2$ is bounded. (not too difficult to show).