# Continuity of function $T: \left( C[0,1],d_{\text{sup}} \right) \rightarrow \left( C[0,1],d_{\text{sup}} \right)$

Let $$T: \left( C[0,1],d_{\text{sup}} \right) \rightarrow \left( C[0,1],d_{\text{sup}} \right)$$

where $$d_{\text{sup}}(f,g) = \mbox{sup}_{x \in [0,1]} |f(x) - g(x)|$$

and the definition of $T$ is: $$T(f)(x) = 2 \cdot f(1-x) - 3$$

Is it true that $T$ is a continuous function?

For me it is really strange example of function. Usually I consider function from real number to real number. But here I have no idea how to begin.

I will grateful for your help.

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what is the definition of $T$ here? –  GA316 Nov 17 '13 at 13:02
I added definition of $T$. –  Thomas Nov 17 '13 at 13:02
Write it as a composition of simpler functions, $T = C \circ B \circ A$, where $A(f)(x) = f(1-x)$; $B(f) = 2\cdot f$, $C(f)(x) = f(x)-3$. Check that $A,B,C$ are continuous. –  Daniel Fischer Nov 17 '13 at 13:23
For me the problem is the same - how to check that $A,B,C$ are continuous? –  Thomas Nov 17 '13 at 13:24
@DanielFischer You check that $T(f)$ is continuous, not $T$! Both need to be done. –  Henno Brandsma Nov 17 '13 at 13:37

For any set $S$ it makes sense to talk about a function $f: S \rightarrow S$, formally this is a set of ordered pairs satisfying some conditions, but a more informal description is that we need for every $s$ in $S$, some unique value $f(s)$ in $S$ as well. In the case at hand, $S$ is a set of (continuous) functions, not a set of numbers, but the principle is the same.
So $T(f)$ must be in the set as well, whenever $f$ is, and you described the rule: as $T(f)$ is itself a function from $[0,1]$ to $\mathbb{R}$, to describe what it is we need to define what $T(f)(x)$ is for every $x$ in $[0,1]$, and it takes the value of $f$ (the input function) at the point $1-x$, multiplies it by 2 and subtracts 3: this is again some real number, uniquely determined when we know $f$, so $T(f)(x)$ is well-defined for all $x \in [0,1]$. Whenever $f$ is continuous between $[0,1]$ and $\mathbb{R}$, so is $T(f)$: we can see $T(f)(x)$ as a composition of continuous functions ($x \rightarrow 1-x, f, t \rightarrow 2t-3$, informally), and so $f$ maps continuous functions to continuous functions as required.
To see continuity in the metric $d_\sup$, you need to compute what $d_\sup(T(f), T(g))$ is, for functions $f,g$. By definition this equals the sup (actually a maximum as we work with continuous functions on a compact space) of all differences $\left|T(f)(x) - T(g)(x)\right|$ where $x$ runs over all values in $[0,1]$. But this equals (using definitions and can calling the $-3$'s) $$|\,2\cdot f(1-x) - 2 \cdot g(1-x) \,| = 2\cdot|\,f(1-x) - g(1-x)\,|$$
and the sup of these over all $x$ is just the same as the sup of $|f(x)-g(x)|$ over all $x$ )if one assumes its max at $a$, the other assumes it at $1-a$ and vice versa).
In short, $d_\sup(T(f), T(g)) = 2d_\sup(f,g)$, and this will easily imply continuity of $T$..