Take the 2-minute tour ×
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.

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....

share|improve this question
    
I'm not sure this works and if it does, how to write it better but here is a thought: –  Rudy the Reindeer Nov 23 '12 at 16:27
    
a) You could show that the image of a bounded set is totally bounded. 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. –  Rudy the Reindeer Nov 23 '12 at 16:30
    
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: $$\int f d\mu \quad \text{versus} \quad \int f\,d\mu.$$ –  kahen Nov 23 '12 at 16:38

2 Answers 2

a) is not compact: it is actually onto, since for any $f$ we have $f=Tf_0$, where $f_0(t)=f(\sqrt t)$.

b) you are right.

c) you are right

d) and e): as Davide says, look at the dimension of the rank of $S_n$.

f) you can write $\int_0^tx(s)ds=\int_0^1x(s)\,1_{[0,t]}(s)ds$ and see that $T$ is compact using A.3 in Appendix A: compact operators

share|improve this answer

a) $T$ is bijective, and $C[0,1]$ is infinite dimensional, so $T$ is not a compact operator.

d) After having showed that $T$ is well-defined, define $S_n(x)(t):=\sum_{j=1}^nx(j^{-1})\frac{t^k}{k!}$. What about the dimension of the rank of $S_n$?

e) The set $\{x_p\colon t\mapsto t^p,p\in\Bbb N\}\subset C[0,1]$ is bounded, and $T(x_p)(t)=e^{t^p}$. The task is to show that no subsequence of $\{T(x_p)\}$ form an equi-continuous set.

f) Arzelà-Ascoli's theorem is useful.

share|improve this answer
    
In d) a finite base for the range of $S_n$ would be $\{ t, t^2, \ldots, t^n \}$. But for e) for example with $n = 2$ we got $x(t) + x(t^2)/2$, which i guess is in some sense an extension of a) and so could not be compact? –  Stefan Nov 23 '12 at 17:45
    
@Stefan I've corrected, as it's actually not the same approach for d) and e). –  Davide Giraudo Nov 23 '12 at 20:48

Your Answer

 
discard

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.