Prove discontinuity of $U(f)= f^3$

Question

Let $U:C([0,1])\to C([0,1])$ defined as $U(f):=f^3$. I want to show it's not continuous in the supremum norm (i.e. $L^\infty$ norm) nor in the $L^1$ norm.

I can show the discontinuity with a counter example, but I couldn't find one. Moreover I know the space $(C^0([a,b]), L^1)$ is not complete, maybe this can help.

I tought to find a sequence of functions that converges in $C([0,1])$ but its cubes don't. Is this the right way?

Answer 1

It is continuous with respect to the supremum norm.

If $f_n\rightarrow f$ in the sup norm, let $M$ be such that $\|f_n\|_\infty,\|f\|_\infty<M$. For any two functions $f,g$ we have $f^3-g^3 = (f-g)(f^2+fg+g^2)$ and $\|fg\|_\infty\leq \|f\|_\infty\cdot\|g\|_\infty$. From this we see that $$\|f^3-f_n^3\|_\infty \leq \|f-f_n\|_\infty\|f^2+f\cdot f_n+f_n^2\|_\infty$$

But $|f^2+f\cdot f_n+f_n^2|\leq 3M^2$ so we have

$$\|f^3-f_n^3\|_\infty \leq 3M^2\|f-f_n\|_\infty\rightarrow 0$$

It is not continuous with respect to the $L^1$ norm

First note that if we can find $M$ such that $\|f_n\|_\infty,\|f\|_\infty<M$ then by using the inequality $\|fg\|_{L^1}\leq \|f\|_{L^1}\cdot \|g\|_\infty$ and arguing as before we will get a convergence.

So in order to find a counter-example we need that no such $M$ exists. Let me first construct a counter-example with non-continuous functions and then explain how one can correct this:

Let $f_n(x)$ be the function which takes the value $n$ on the inverval $[0,\frac{1}{n^2}]$ and zero otherwise then

$$\int_0^1 |f_n(x)|dx = \int_0^{\frac{1}{n^2}}n = n\cdot \frac{1}{n^2}=\frac{1}{n}\rightarrow 0$$ therefore $f_n\rightarrow 0$ in $L^1$ (and is not bounded of course).

However $(f_n(x))^3$ takes the value $n^3$ again on $[0,\frac{1}{n^2}]$. Same calculation as before gives

$$\int_0^1 |f_n(x)|dx = \int_0^{\frac{1}{n^2}}n = n^3\cdot \frac{1}{n^2}=n\rightarrow \infty$$

which diverge!

In order to deal with the discontinuity one can replace $f_n$ on a small interval $[\frac{1}{n^2},\frac{1}{n^2}+\varepsilon]$ by a "line" which connectes $f_n(\frac{1}{n^2})$ with $0$ (try to draw it) and then argue as before.

Answer 2

Considering the $L^1$-norm, one might think of functions like $f_n=1_{[0,n^{-2}]}\cdot n$. They converge to zero with respect to the $L^1$-norm, but the cubes diverge. However, these functions are obviously not continuous. This is not a major problem though, you just have to "smoothen" the jump. Do you know how to do that?