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I want to examine which of the following operators $T \colon C[0,1] \to C[0,1]$. are compact, by some I think I got the argument, but others I have no idea.

a) $Tx(t) = x(t^2)$

Guess it is compact, but I have no idea how to proof this?

b) $Tx(t) = x(0) + tx(1)$

Here the range of $T$ consist of lines, i.e. the set $\{ n + m \cdot x : n,m \in \mathbb{R} \}$, this set is finite-dimensional because $\{ \mathbb{1}, \operatorname{id} \}$ are a base ($1$ denotes the constant function $1(x) = 1$ for all $x$).

c) $Tx(t) = \int_0^1 e^{st} x(s) \mathrm{d}s$

This is compact according to example A.2 from Appendix A: Compact Operators

d) $Tx(t) = \sum_{k=1}^{\infty} x(\frac{1}{k}) \frac{t^k}{k!}$

Guess here I could use arguments similar to those

How to prove that an operator is compact?

Proof that operator is compact

because $x(\frac{1}{x})$ is bounded on $[0,1]$ and the series $\sum_{k=1}^{\infty} \frac{t^k}{k!}$ converges to $e^t - 1$.

e) $Tx(t) = \sum_{k=0}^{\infty} \frac{x(t^k)}{k!}$.

Here I have no idea how to proof or disproof compactness of $T$?

f) $Tx(t) = \int_0^t x(s) \mathrm{d} s$

Here I have no glue too....

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    I'm not sure this works and if it does, how to write it better but here is a thought:2012-11-23
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    a) You could show that the image of a bounded set is [totally bounded](http://en.wikipedia.org/wiki/Compact_operator#Equivalent_formulations). Let $X \subset C[0,1]$ be a bounded set that is: $\sup_{f \in X} \|f\|_\infty = K < \infty$. The map $t \mapsto t^2$ is continuous and its image is $[0,1]$. Hence if $X$ is bounded by $K$ then $TX$ is still bounded by $K$ since the domain stays the same. I'm not sure how to finish. One has to give a finite cover of $TX$ of [sets of fixed size](http://en.wikipedia.org/wiki/Totally_bounded_space).2012-11-23
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    Small MathJax/LaTeX tip: I find that integrals look better when you put a thin space (`\,` in math mode) in front of the 'd'. Compare these two: $$\int f d\mu \quad \text{versus} \quad \int f\,d\mu.$$2012-11-23

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