Is $\Phi:L_p\rightarrow L_p$ continuous?

Question

Following this question: Is $\Phi:L_p(X,\mathcal{A},\mu)\rightarrow L_p(X,\mathcal{A},\mu)$ a function well defined?

The answer given is that if there exists a constant $C>0$ such that $$\|\varphi\circ f\|_p\leq C\|f\|_p \,(1)$$ then you get that the function $\Phi$ is defined and continuous. I get why it's defined, but I don't see why this implies $\Phi$ is continuos. I know that if we add that $\varphi$ is lineal; that is, $$\varphi(x+y)=\varphi(x)+\varphi(y)\,(2)$$ then you get that $\Phi$ is a linear operator and the condition (1) implies that $\Phi$ is bounded hence continuous, but without that extra condition (the extra conditions is (2)) I don't see why would be $\Phi$ continuos.

Answer 1

$\phi$ does not need to be linear.

$\Phi(f)=\phi\circ f$ such that $\phi:\mathbb{R}\rightarrow\mathbb{R}$ Lipschitz and $\phi(0)=0$ would do the job.

Indeed if $\phi$ is Lipschitz with constant $c$

$|\phi\circ f)|=|\phi\circ f)-\phi(0)|\leq c|f|\in L_p$

and

$\|\phi\circ f-\phi\circ g\|_p\leq c\|f-g\|_p$

The solution you are referring too assume Lipschitz, but that alone does not guarantee that $\phi\circ f\in L_p$ for $f\in L_p$.

Answer 2

Perhaps you need this:

Monotonicity of $\ell_p$ norm which shows that there is, indeed, such a $C.$