So here is my question,

I had to show that the following operator is compact,

$$T:C[0,1]\rightarrow C[0,1]$$ $$f\mapsto\int_0^tf(s)ds$$ with $||f||=\mathrm{sup}_{x\in[0,1]}|f(x)|$

I think I managed to prove it using the "bounded sequence definition" of compactness.

As I saw in many posts a common way to prove that an operator is compact, is finding a squence of finite ranked operators which converges to the operator. I was wondering if this is possible for the upper one? I know that there exists a sequence for sure but I am not sure if it is possible to write it down explictly. Can somebody help me?

Thanks in advance.

  • $\begingroup$ What is the metric on your space? $\endgroup$
    – 5xum
    May 27, 2014 at 11:16
  • $\begingroup$ @5xum sorry, the supremum norm. $\endgroup$
    – Thorben
    May 27, 2014 at 11:20
  • 1
    $\begingroup$ I do not know the answer, but the closure of finite rank operator might not be the set of compact operator. This is true when working on Hilbert space but not general Banach spaces. $\endgroup$
    – user99914
    May 27, 2014 at 11:23
  • 2
    $\begingroup$ @John Can you give an example of a Banach space where there exists a compact operator that isn't the norm-limit of finite-rank operators? $\endgroup$ May 27, 2014 at 12:10
  • $\begingroup$ @DanielFischer: I do not know. That seems to be a hard problem. I only can refer you to the wiki articles en.wikipedia.org/wiki/Approximation_property $\endgroup$
    – user99914
    May 28, 2014 at 2:03

1 Answer 1


For $n\in\mathbb{Z}^+$, let $(h_{n,k})_{1 \leqslant k \leqslant n}$ be a continuous partition of unity,

$$h_{n,k}(x) = \begin{cases}\qquad 0 &, x < \frac{k-1}{n} - \frac{1}{2^{n+1}}\\ \frac{1}{2} + 2^n\left(x-\frac{k-1}{n}\right) &, \frac{k-1}{n} - \frac{1}{2^{n+1}} \leqslant x \leqslant \frac{k-1}{n}+\frac{1}{2^{n+1}}\\ \qquad 1 &, \frac{k-1}{n}+\frac{1}{2^{n+1}} < x < \frac{k}{n} - \frac{1}{2^{n+1}}\\ \frac{1}{2} - 2^n\left(x-\frac{k}{n}\right) &, \frac{k}{n} - \frac{1}{2^{n+1}} \leqslant x \leqslant \frac{k}{n} + \frac{1}{2^{n+1}}\\ \qquad 0 &, \frac{k}{n} + \frac{1}{2^{n+1}} < x\end{cases}$$

for $1 < k < n$, and the interval where $h_{n,k}(x) = 1$ extending to $0$ resp. $1$ for $k = 1$ resp. $k = n$.

Then we can define a sequence of approximating projections with finite rank,

$$P_n(f) = \sum_{k=1}^n f\left(\frac{k-\frac{1}{2}}{n}\right)\cdot h_{n,k},$$


$$T_n(f)(t) = \int_0^t P_n(f)(s)\,ds$$

is a sequence of finite-rank operators approximating $T$.

Another way to use the partition of unity to obtain a sequence of approximating finite-rank operators is to set

$$\tilde{T}_n(f) = \sum_{k=1}^n\int_0^{\frac{k-\frac{1}{2}}{n}} f(s)\,ds\cdot h_{n,k}.$$

  • $\begingroup$ Hi Fischer, how do you come up with such a partition of unity? I mean, it seems hard to write down by bare hand. Can you tell me something like motivation behind it? Thanks. $\endgroup$
    – Sam Wong
    Jun 27, 2018 at 13:01
  • $\begingroup$ The motivation is that one can uniformly approximate a continuous function on $[0,1]$ by step functions, which is familiar from the Riemann integral. But step functions are generally not continuous, so one needs to modify the characteristic functions of the subintervals to get a continuous partition of unity. $\endgroup$ Jun 27, 2018 at 13:19
  • $\begingroup$ I got the point. Thanks! $\endgroup$
    – Sam Wong
    Jun 27, 2018 at 14:45

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