Bounded linear transformation 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.


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*Prove that $T$ is bounded. What is the value of $\|T\|$?

*Give a necessary and sufficient condition on $g$ for $T$ to be onto.

*Give a necessary and sufficient condition on $g$ for $T$ to be one-to-one.

*Give a necessary and sufficient condition on $g$ for $T$ to be an isometry.

 A: *

*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 $\|T\| \geq 1$.

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*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 $T$. 

*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. 


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

*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.


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*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.

*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 $g(X) = Y$.
