Consider a function $f:\mathbb R\to \mathbb R$, and consider it as a vector in the space of functions from $\mathbb R\to\mathbb R$.
Just as how a linear operator on a finite $n$-dimensional vector space $\mathbb R^n$ can be represented by a $n$-$n$ matrix, i.e. an element of $\mathbb R^{n\times n}$, we could consider a linear operator on $\mathbb R^\mathbb R$ to be represented by an $R$ by $R$ function, i.e. an element of $\mathbb R^{\mathbb R\times \mathbb R}=\mathbb R^{\mathbb R^2}$.
So we could consider a (differentiable) function $f:\mathbb R\to \mathbb R$, and a “continuous matrix” $D:\mathbb R^2\to \mathbb R$.
The analogous operation to “matrix multiplication” would be:
$$Df(x)=\int_{-\infty}^\infty D(x,y)f(y)dy$$
Linear operators of this form are called “linear integral operators”. It seems to me that no such $D(x,y)$ exists that would make $Df(x)$ be the derivative of $f(x)$. But perhaps something like a generalized function $D(x,y)$, analogous to the dirac-delta function?
Am I right that no $D(x,y)$ exists that represents the derivative operator (i.e. the derivative cannot be represented as a linear integral operator)? What are the implications of this? I have almost no knowledge of functional analysis, and I’m trying to get a better idea.
How far can we go with representing linear operators as “infinite dimensional matrices” like this? How powerful is this approach? It seems to me that it is not very powerful, and I’m trying to understand what functional analysists do to overcome this.
Intuitions about finite dimensional vector spaces/linear algebra, don’t always transfer to infinite-dimensional vector spaces/functional analysis. Is the fact that not all linear operators are linear integral operators the “reason” for this? I.e. do properties of finite dimensional linear operators transfer to linear integral operators?
