The product rule on $C[0, 1]$ for different norms

Question

If $f_n → f$ and $g_n → g$, does $f_n g_n → fg$ in the space $C[0, 1]$ for the norms $||.||_1$ and $||.||_∞$

Give a proof or counterexample for each.

I know that $||.||_1$ is the sum of the magnitudes of the values, and $||.||_∞$ is the biggest magnitude of the values. From this I assume $||.||_1$ DOES work whilst $||.||_∞$ doesn't, but I am unsure on if this is true, and if so, how to show it.

Thanks in advance

Answer 1

Unfortunately, both of your guesses are incorrect.

To see the $\|\,\|_1$ result fails, let

$$f_n(x)= \frac{1}{[\ln (n+1)(x+1/n)]^{1/2}}.$$

Verify that $f_n\to 0$ in the $\|\,\|_1$-norm. However $f_n\cdot f_n$ does not converge to $0\cdot 0=0$ in the $\|\,\|_1$-norm, as you can check.

The $\|\,\|_\infty$ result is true. This is classic and is easier to show. We have $f_n\to f,g_n\to g$ uniformly on $[0,1].$ This implies there is a uniform bound on all of these functions. Now use

$$f_ng_n-fg= f_ng_n-fg_n + fg_n-fg$$

to see $f_ng_n\to fg$ uniformly.

Answer 2

In my earlier answer, I've addressed only the uniform norm; as, @zhw. noted, I sloppily missed the $\|.\|_1$-norm (shame on me) which zhw. did solve. And still, to make up for my misdeed, let me present another example which shows that $\|.\|_1$-norm is not continuously multiplicative.

Let $\,a>0\,$ and $\,0<b\le 1\,$ be two otherwise arbitrary reals. Define:

$$ \forall_{x\in[0;1]}\quad F_{a\,b}(x) \ :=\ \max(a\!\cdot\!(b-x)\ \ 0) $$

hence $$ \forall_{x\in[0;1]}\quad F^2_{a\,b}(x)\ = \ \min(a^2\!\cdot\!(b-x)^2\ \ 0) $$

so that

$$ \|F_{a\ b}\|_{_1}\ =\ \frac 12\cdot a\cdot b^2 $$

and

$$ \|F^2_{a\ b}\|_{_1}\ =\ \frac 13\cdot a^2\cdot b^3 $$

Define

$$ \forall_{n\in\Bbb N}\quad f_n := F_{_{n^5\,\ n^{-3}}} $$

Then

$$ \|f_n\|_{_1} = \frac 1{2\cdot n}\quad\to\quad 0 $$ while $$ \|f^2_n\|_{_1} = \frac n3\quad\to\quad\infty $$

This shows the mentioned lack of continuity.   Great!

Answer 3

For a counterexample to the $\|\cdot\|_1$ norm, consider $$f_n(t) = \begin{cases} n-n^3t, &\text{ if } t \in \left[0,\frac1{n^2}\right]\\ 0, &\text{ if } t \in \left[\frac1{n^2},1\right]\\ \end{cases}$$

Then $\|f_n\|_1 = \frac1{2n} \to 0$ but $\|f_n^2\|_1 = \frac13 \not\to 0$.