# Is the product rule $f_n \to f$, $g_n\to g \Rightarrow f_ng_n\to fg$ true in the space $C[0,1]$?

Is the product rule $$f_n \to f$$, $$g_n\to g \Rightarrow f_ng_n\to fg$$ true in the space $$C[0,1]$$? The answer depends on the norm. Give a proof, or give a counterexample, for the norms $$||\cdot||_1$$ and $$||\cdot||_{\infty}$$.

My attempt:

We need to show that $$||f_ng_n-fg||_{\infty}\to 0$$:

$$||f_ng_n-fg||_{\infty}=||f_ng_n-f_ng+fg_n-fg||_{\infty}\leq n||f_n-g||_{\infty}+n||g_n-g||_{\infty}$$

Since $$f_n\to f$$ and $$g_n\to g$$ uniformly:

$$n||f_n-f||_{\infty}+n||g_n-g||_{\infty}\to 0$$

Hence, $$f_ng_n\to fg$$ uniformly and the product rule works

Would this be correct?

• Note that $C[0,1]$ is a Banach algebra. Your result follows from the fact that the multiplication on a Banach algebra is (jointly) continuous. May 10 '20 at 23:55
• @Calculix: the fact that multiplication is continuos is exactly what the OP is asked to prove. May 15 '20 at 18:32

## 1 Answer

No it is not correct. How do you know that $$n\Vert g_n - g\Vert_\infty\to 0$$? I also don't see why the estimation where you introduce $$n$$ should hold.

Rather, use that (why does this inequality hold?)$$\Vert f g \Vert_\infty \le \Vert f \Vert_\infty \Vert g \Vert_\infty$$

to deduce

$$\Vert f_ng_n -fg \Vert_\infty = \Vert f_n(g_n-g) + g(f_n-f)\Vert_\infty$$ $$\leq \Vert f_n \Vert_\infty \Vert g_n-g \Vert_\infty + \Vert g \Vert_\infty \Vert f_n-f\Vert_\infty$$

together with the fact that $$\{\Vert f_n \Vert_\infty\}_n$$ is a bounded sequence (why is this true and why is this relevant?)

I do not supply a hint for $$\Vert \cdot \Vert_1$$ since you did not include your attempt for that subquestion.