Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $T$..

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.