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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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  • $\begingroup$ I'm not sure this works and if it does, how to write it better but here is a thought: $\endgroup$ Commented Nov 23, 2012 at 16:27
  • $\begingroup$ 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. $\endgroup$ Commented Nov 23, 2012 at 16:30
  • $\begingroup$ 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: $$\int f d\mu \quad \text{versus} \quad \int f\,d\mu.$$ $\endgroup$
    – kahen
    Commented Nov 23, 2012 at 16:38

2 Answers 2

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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

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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.

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  • $\begingroup$ 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? $\endgroup$
    – StefanH
    Commented Nov 23, 2012 at 17:45
  • $\begingroup$ @Stefan I've corrected, as it's actually not the same approach for d) and e). $\endgroup$ Commented Nov 23, 2012 at 20:48

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