how to prove this operator to be compact

Question

Let $X=C(I)$, where $I=[0,1]$. Let $T$ be a Volterra-type integral operator:

$$T(f)(x)=\int_0^xf(t)dt, f\in X$$

Show that, as an operator $X\rightarrow X$, $T$ is compact I dont know how to show T is compact, could you plz help me

Answer 1

Here is a direct approach, that uses no deep theorem, other than the fact that limits of finite-rank operators are compact.

Since we are integrating continuous functions, it is enough to use regularly spaced partitions. For each partition $P=0,1/n, 2/n,\ldots,1$ we define functions $$ \Delta_{n,k}(x)=\begin{cases} 1/n,&\ x\leq k/n\\ \ \\ 0,&\ x>k/n\end{cases} $$ Then the operators $R_{n,k}$ given by $$ R_{n,k}f=f(k/n)\,\Delta_{n,k} $$ are rank-one, and $$\tag1 \int_0^xf(t)\,dt=\lim_{n\to\infty}\sum_{k=1}^n R_{n,k} f\,(x) $$ is a limit of finite-rank operators. All that remains is to show that the limit is not only pointwise, but uniform. Let $\varepsilon>0$. Since $f$ is uniformly continuous, there exists $\delta>0$ such that $|f(t)-f(s)|<\varepsilon$ if $|t-s|<\delta$. So, if $n>1/\delta$, \begin{align} \left|\int_0^xf(t)\,dt-\sum_{k=1}^nR_{n,k}f\,(x) \right| &=\left|\sum_{k=1}^{\lfloor nx\rfloor}\int_{k/n}^{(k+1)/n}(f(t)-f(k/n))\,dt \right|\\ \ \\ &\leq\sum_{k=1}^{\lfloor nx\rfloor}\int_{k/n}^{(k+1)/n}\varepsilon\,dt \\ \ \\ &\leq\sum_{k=1}^{n}\int_{k/n}^{(k+1)/n}\varepsilon\,dt \\ \ \\ &=\varepsilon\,\sum_{k=1}^n\frac1n=\varepsilon. \end{align} So the limit in $(1)$ is actually a limit in the uniform norm, and so $T$ is a limit of finite-rank operators, which makes it compact.

Answer 2

I assume we're using

$\Vert f(x) \Vert = \sup_{x \in I} \vert f(x) \vert \tag 1$

as the norm on $C(I)$; then $C(I)$ is Banach, as is well known.

Let $f_n(x)$, $n \in \Bbb N$, be a sequence in $C(I)$ such that

$\Vert f_n(x) \Vert \le M \tag 2$

for all $n$; that is $f_n(x)$ is a bounded sequence. We have, for any $f(x) \in C(I)$,

$\vert Tf(x) \vert = \vert \displaystyle \int_0^x f(t) \; dt \vert \le \int_0^x \vert f(t) \vert \; dt \le \int_0^x \Vert f(x) \Vert \; dt \le \Vert f(x)\Vert \int_0^1 dt = \Vert f(x) \Vert \tag 3$

whence

$\Vert Tf(x) \Vert \le \Vert f(x) \Vert; \tag 4$

thus, by (2),

$\Vert Tf_n(x) \Vert \le M, \tag 5$

which shows the sequence $Tf_n(x)$ is uniformly bounded. Also, for any $f(x) \in C(I)$ and $y > x$,

$\vert Tf(y) - Tf(x) \vert = \vert \displaystyle \int_0^y f(t) \; dt - \int_0^x f(t) \; dt \vert = \vert \int_x^y f(t) \; dt \vert \le \int_x^y \vert f(t) \vert \; dt$ $\le \displaystyle \int_x^y \Vert f(x) \Vert \; dt = \Vert f(x) \Vert \int_x^y dt = \Vert f(x) \Vert (y - x); \tag 6$

a similar calculation with $y < x$ shows that

$\vert Tf(y) - Tf(x) \vert = \Vert f(x) \Vert \int_x^y dt = \Vert f(x) \Vert (x - y); \tag 7$

together (6) and (7) imply

$\vert Tf(y) - Tf(x) \vert = \Vert f(x) \Vert \int_x^y dt = \Vert f(x) \Vert \vert x - y \vert, \tag 8$

for all $x, y \in I$, so we have in fact

$\vert Tf_n(y) - Tf_n(x) \vert = \Vert f_n(x) \Vert \vert x - y \vert \le M \vert x - y \vert; \tag 9$

now for any $\epsilon > 0$ we pick $\delta = \epsilon / M$ and find that, for $\vert x - y \vert < \delta$,

$\vert Tf_n(y) - Tf_n(x) \vert = M \vert x - y \vert < M \delta = M(\dfrac{\epsilon}{M}) = \epsilon; \tag{10}$

this result is independent of $n$; thus the sequence $Tf_n(x)$ is equicontinuous. It now follows from the Arzela-Ascoli theorem that $Tf_n(x)$ has a uniformly convergent subsequence, which means the subsequence converges in the norm $\Vert \cdot \Vert$ on $C(I)$. Thus $T$ meets the definition (well, one of the equivalent definitions) of a compact operator, and is thus compact.