Compact operator as a limit of finite ranked operators

Question

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.

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,

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

and

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

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

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