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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?

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  • $\begingroup$ 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. $\endgroup$
    – Calculix
    May 10 '20 at 23:55
  • $\begingroup$ @Calculix: the fact that multiplication is continuos is exactly what the OP is asked to prove. $\endgroup$ May 15 '20 at 18:32
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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.

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