Point of proving operators are bounded

Question

I've been learning a bit of analysis and a lot of what I've been reading about deals with proving certain operators are $L^p$ bounded, but I haven't seen applications yet to proving that these operators are bounded. So my question is: why do analysts care so much about proving certain operators are bounded?

Answer 1

Here's a nice example: you may have seen the result that if the sequence of functions $f_n:[0,1] \to \Bbb R$ converges uniformly to $f$, then the functions $$ g_n(t) = \int_0^t f_n(t)\,dt $$ will converge uniformly to their limit $g(t) = \int_0^t f(t)\,dt$. This is a very handy result for computing certain integrals.

Another way to state this result is to state that the linear operator $S:L^{\infty}[0,1] \to L^\infty[0,1]$ given by $$ [S(f)](t) = \int_0^t f(t)\,dt $$ is bounded, and therefore continuous.

In contrast, we cannot conclude that the same will hold for $f_n'$, even if the $f_n$ are each differentiable. The "differentiation operator" fails to be bounded, which can make life difficult.

Answer 2

One can prove bounded linear operator is equivalent to continuous linear operator. In short, analysts always want their linear "functions" to be continuous.