Problem: Fix a continuous function $\phi:[0,1]\to[0,1]$ and define the operator $C_\phi:C([0,1])\to C([0,1])$ by $(C_\phi f)(x)=f(\phi(x)).$ Prove that this is a bounded linear operator with $\|C_\phi\|\le1$.
Attempt: Consider $\|C_\phi\|_{op}$, now,
$$\|C_\phi\|_{op}=\sup\{\|C_\phi f\|_\infty:f\in C([0,1])\}=\sup\{\sup\{|(C_\phi f)(x)|:x\in[0,1]\}:f\in C([0,1])\}$$
Regard the "inner supremum" on it's own for the time being. Using the definition of $C_\phi$ we see that,
$$\sup\{|(C_\phi f)(x)|:x\in[0,1]\}=\{\sup\{|f(\phi(x))|:x\in[0,1]\}$$
Since $\phi:[0,1]\to[0,1]$ it is clear that $\phi(x)=y$ where $y\in[0,1]$, we get that,
$$\sup\{|f(\phi(x))|:x\in[0,1]\}=\sup\{|f(y)|:y\in[0,1]\}=\|f\|_\infty$$
So that, going back to the "sup of the sup" we had before,
$$\sup\{\sup\{|(C_\phi f)(x)|:x\in[0,1]\}:f\in C([0,1])\}=\sup\{\|f\|_\infty:f\in C([0,1])\}=\|f\|_\infty$$
But then, putting this all together, we have that $\|C_\phi\|_{op}=\|f\|_\infty$, and I don't believe that this is right. I think that it is, rather, an incorrect means to approach the problem.
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On the other hand if we just considered,
$$\|C_\phi f\|_\infty=\sup\{|(C_\phi f)(x)|:x\in[0,1]\}=\sup\{|f(y)|:y\in[0,1]\}=\|f\|_\infty$$
So that in particular, $\|C_\phi f\|_\infty=\|f\|_\infty$ which would mean that $\|C_\phi\|_{op}=1$. Now saying that something is equal to something does still satisfy saying that it is less than or equal to something. So I have to wonder, is this sufficient to answer the problem? Or have I done something wrong, this second time round, which has made it seem easier than it should?
