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For a compact metric space $X$, $C(X)$ denotes the space of continuous real-valued functions on $X$ equipped with the supremum norm. Let $X$ and $Y$ be compact metric space and let $g:X \to Y$ be a continuous map. Define $T: C(Y) \to C(X)$ by $T(f) = f\circ g$ . Clearly, $T$ is a linear transformation.

  1. Prove that $T$ is bounded. What is the value of $\|T\|$?
  2. Give a necessary and sufficient condition on $g$ for $T$ to be onto.
  3. Give a necessary and sufficient condition on $g$ for $T$ to be one-to-one.
  4. Give a necessary and sufficient condition on $g$ for $T$ to be an isometry.
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    What is your question? What did you try and where did you get stuck?2012-03-21
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    Hey @Chung, why don't you tell us where you're stuck and let us help you do this homework? I'm sorry you got 4 downvotes, I don't think that's deserved. The downvoters might remove the downvotes if you post your attempts and show us where you got stuck.2012-03-24
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    Why don`t you give a hints instead of him giving what he tried? Maybe some hints will be useful here.2012-03-25
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    It would be nice if people asked for help rather than just demanding answers...2012-03-25
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    I feel @t.b. 's comment here http://math.stackexchange.com/questions/123133/hahn-banach-theorem#comment284673_123133 still applies2012-03-25
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    More to the point: as a teacher, being told "Sir I am stuck" is next to useless. Being told "I don't understand what I have to prove" or "I can see how to do this in a special case, like this, but I don't know how to generalize" allows some actual *education* to be attempted2012-03-25
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    Are you @Danny?2012-03-27
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    @MattN. I'm afraid I disagree, and writing out a detailed answer as you have done below seems (to me) wasted effort until the OP takes note of http://meta.math.stackexchange.com/questions/1803/how-to-ask-a-homework-question2012-03-27

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  1. Note that $\|T(f)\|_{C(X)} = \sup_{x \in X}\left | f \circ g (x) \right | = \sup_{y \in g(X)}\left | f (y) \right | \leq \sup_{y \in Y}\left | f (y) \right | = \| f \|_{C(Y)}$ hence $\frac{\|T(f)\|_{C(X)}}{\| f \|_{C(Y)}} \leq 1$ for $f \neq 0$ and hence $\|T\| \leq 1$.

    On the other hand, note that for $f(x) = 1$ you have $\|T(f)\|_{C(X)} = 1$ and hence $\|T\| \geq 1$.

    1. A necessary condition is that $g$ is injective. To see this assume that $g$ is not injective. Then there are $x, x^\prime$ such that $g(x) = g(x^\prime)$. So $T$ cannot map to functions $h: X \to \mathbb R$ with $h(x) \neq h(x^\prime)$. But such functions exist: since $X$ is a metric space you can use Urysohn's lemma to get a function $h$ such that $h(x) = 1$ and $h(x^\prime) = 0$. This $h$ is not in the image of $T$.

    2. This is also sufficient. To see this assume that $g$ is injective. Then $g$ is a bijection $g: X \to g(X)$ hence has a continuous inverse $g^{-1} : g(X) \to X$ because $X$ is compact and $g(X)$ is Hausdorff. Now let $h: X \to \mathbb R$ be any function in $C(X)$. Note that $f := h \circ g^{-1}$ is a function from the subset $g(X)$ of $Y$ to $\mathbb{R}$ with the property that $T(f) = f \circ g = h$. Now you can extend $f$ to all of $X$ using Tietze's extension theorem since $g(X)$ is compact and hence closed.

    1. To find a sufficient condition for $T$ to be injective note that if $g$ is surjective, $f(g(x)) = f^\prime (g(x))$ implies $f = f^\prime$.

    2. Surjectivity of $g$ is also necessary: Assume $g$ is not surjective then there exists a $y$ in $Y \setminus g(X)$. Let $f \in C(X)$. Note that $g(X)$ is closed in $Y$ and use Urysohn again to get a function $f^{\prime \prime} : X \to \mathbb{R}$ with $f^{\prime \prime}\mid_{g(X)} = 0$ and $f^{\prime \prime} (y) = 1$. Then define $f^\prime := f + f^{\prime \prime}$. Now $f^\prime$ is a function in $C(X)$ such that $f \neq f^\prime$ and $T(f) = T(f^\prime)$ hence $T$ is not injective.

    1. Note that for $T$ to be an isometry it suffices to show $\| f \|_\infty = \|T(f) \|_{C(Y)}$ by linearity of $T$. This holds if $g$ is surjective so we have a sufficient condition. Also note that $\|T(f)\| = \|f\circ g\| = \|f\|$ is the same as saying $\sup_{y \in Y} | f (y) | = \sup_{y \in g(X)} | f (y) | $ which holds if $g(X)$ is dense in $Y$, so we have found another sufficient condition. Note that if $g(X)$ is dense in $Y$ then $g(X) = Y$ because $X$ is compact and hence $g(X)$ is closed.

    2. Now assume that $g(X)$ is not dense in $Y$. Then we can find a $y$ in $Y$ and an $\varepsilon > 0$ such that $B(y, \varepsilon) \cap g(X) = \varnothing$. Define $$f(x) := \begin{cases} \frac{\varepsilon - d(x,y)}{\varepsilon} & x \in B(y,\varepsilon) \\ 0 & \text{ otherwise} \end{cases}$$ Then this is a function that is zero everywhere except inside an epsilon ball contained in $Y \setminus g(X)$. Hence we have found an $f$ such that $\sup_{y \in Y} | f (y) | \neq \sup_{y \in g(X)} | f (y) | $ and hence $T$ is not an isometry.

    So $T$ is an isometry if and only if $g(X) = Y$.

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    In a metric space, Urysohn is usually overkill...2012-03-27
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    @Yemon: Urysohn is straightforward for metric spaces: let $A, B$ be disjoint and closed and non-empty. Put $f(x) = \frac{d(x,A)}{d(x,A)+d(x,B)}$. So either you call that Urysohn or it is indeed overkill :)2012-03-27
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    @t.b. If we call that Urysohn, then I am calling the connectedness of the circle the Kadison-Kaplansky conjecture...2012-03-27
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    @Yemon: Apart from the fact that this is an infinitely silly debate, [Urysohn gave the formula himself](http://i.stack.imgur.com/2jQWW.png) as a footnote to [his article](http://dx.doi.org/10.1007/BF01208659).2012-03-27
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    @t.b. Touché. But in the place where Matt N. appeals to Urysohn, one merely needs a continuous function that is 1 on one point and 0 at the other. This really, really does not need anything as complicated as Urysohn, metric or otherwise2012-03-27
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    In the first part, Can you please explain how did you get that the norm is bigger or equal 1. I also don`t understand how do you apply the theorems that you mentioned in the answer and how do you guarantee that the conditions of the theorems are satisfied like the space being a normal space. Can you please give some details. Thanks.2012-03-27