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?
Best 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.