An explicit example of a compact integral operator in one dimension to help with intuition? 
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*I am about to start studying compact operators and I always find the best way to get intuition on a new area is to start with a simple example. So what would be a an example of a compact integral operator, lets call it $T$, in one dimension?

*To further help with intuition, how could we 'modify' the operator $T$ to make it not compact? ..by this I mean, we know that the closed interval $[0,1]$ is complete, however if we removed the endpoints and are left with the open interval $(0,1)$ which is not complete. Is there some analogous way of modifying the one dimensional compact integral operator $T$ such that it 'looks' almost the same as the original operator but it is no longer compact?

*I have heard that compact operators have special spectral properties. It would be great if a spectral representation/decomposition of the operator $T$ could be specified.

 A: First of all, $L^2[0,1]$ and $L^2(0,1)$  are the same as Hilbert spaces, since $\{0,1\}$ has Lebesgue measure zero. So, you should not really expect anything which happens at those points to have any effect (Remark: The preceding claim is a bit problematic if we want to deal with unbounded operators, so that a domain needs to be given. Then, how the functions in our domain behave at the endpoints of the interval can come into play). 
By an "integral operator" on $L^2[0,1]$, let us understand an operator $T$ which is given by $(Tw)(x)=\int_0^1 K(x,y) w(y) \ dy$ where $K : [0,1]^2 \to \mathbb{R}$ is some reasonably well-behaved function. We say that $K$ is the integral kernel for the operator $T$. The idea is that $K$ is something like a $[0,1]$ by $[0,1]$ matrix for the operator $T$. Just how well-behaved $K$ needs to be in order for $T$ to be a bounded/compact/whatever operator turns out to be a rather delicate issue.
One way, however, to immediately convince yourself that there is some connection between operators given by integral kernels and compact operators is to consider the case of finite rank operators. Let $H$ be some Hilbert space with inner product $\langle \cdot , \cdot \rangle$ linear in the first slot. A unit vector $v$ defines a rank-1 projection operator $P_v$  by the formula:
$$P_v(w) = \langle w,v \rangle v.$$
A little more generally, a pair of unit vectors $u,v \in H$ defines a rank-1 partial isometry by the formula
$$S_{u,v}(w) = \langle w,v \rangle u.$$
Finally, every finite rank operator $F:H \to H$ can be written as
$$F = \sum_{i=1}^n \lambda_i S_{u_i,v_i}$$
where $u_1,\ldots,u_n$ is an orthonormal sequence in $H$, $v_1,\ldots,v_n$ is another orthonormal sequence in $H$, and $\lambda_1,\ldots, \lambda_n$ are positive constants. This is basically polar decomposition for a finite-rank operator. If we want to get all compact operators, we have to let $n=\infty$, and also demand $\lambda_i \to 0$ as $i \to \infty$. But, for purposes of illustration, it is entirely sufficient for us to consider finite-rank operators.
Now let's consider the case where $H=L^2[0,1]$ and the inner product is
$$\langle u,v \rangle = \int_0^1 u(x) \overline v(x) \ dx$$
Then, the projection $P_v$ is given by
$$(P_v(w))(x) = \langle w, v \rangle v(x) = \int_0^1 w(y) \overline v(y) v(x) \ dy = \int_0^1 K_v (x,y) w(y) \ dy$$
where
$$K_v(x,y) = v(x) \overline v(y)$$
Similarly, the partial isometry $S_{u,v}$ is given by
$$(S_{u,v}(w))(x) = \int_0^1 K_{u,v}(x,y) w(y) \ dy$$
where
$$K_{u,v}(x,y) = u(x) \overline v(y).$$
More generally still, the generic finite rank operator $F$ indicated above is given by
$$(Fw)(x)=\int_0^1 K(x,y) w(y) \ dy$$
where 
$$K(x,y) =\sum_{i=1}^n \lambda_i u_i(x) v_i(y).$$
The conclusion is that, in a rather direct way, we can think of finite rank operators as coming from integral kernels. Since compact operators are (norm) limits of finite rank operators, it is reasonable to think that compact operators should be given by similar integral kernels where we just let the sum run to infinity. The only issue left, and it is not a trivial one, is to decide what exactly we mean when we talk about convergence of kernels.
