Consider the integral operator $T : C([0,1])\to C([0,1])$ given by
$$Tf(t)=\int_0^1 K(t,\tau)f(\tau)d\tau.$$
I'm solving one exercise which is to show this operator is bounded. The exercise is from a mathematical physics course.
Also, I want to show this without any results from Lebesgue integration. So I'm disconsidering here theorems like the dominated convergence theorem.
My first idea was to consider the supremum norm. Indeed if we were considering it we would have:
$$\left|\int_0^1 K(t,\tau)f(\tau)d\tau\right|\leq \|K(t,\cdot)f\|_{\infty}$$
Now, $f$ is continuous and $[0,1]$ is compact so that $f$ has a maximum. But we have that $t$ dependence on $K$.
Also, although the exercise doesn't say anything about the norm or $K$, I believe the norm being considered here is the one which comes from the usual inner product $\langle, \rangle$ given by
$$\langle f,g\rangle = \int_0^1 f(t)g(t)dt.$$
I've been thinking about this for some time now and I didn't get any further. How can I show this operator is bounded?
Best Answer
Dominated convergence and related theorems don't apply here anyway.
The norm here is almost certainly the supremum norm since this is the norm that makes $C([0,1])$ a complete space. Also, you'll need to to know something about $K$ in order to do this. For example, if $K$ is continuous the following argument works: what you need to do is find a $C > 0$ so that $$\| Tf \|_\infty \le C \| f \|_\infty$$ for all $f \in C([0,1])$ where $\| f \|_\infty = \sup_{t \in [0,1]} \lvert f(t) \rvert$. If $K$ is continuous on $[0,1] \times [0,1]$, then by compactness, $K$ is bounded; say $\lvert K(t,\tau) \rvert \le M$ for all $(t,\tau) \in [0,1]\times [0,1]$. Then we see that for any $f \in C([0,1])$, $t \in [0,1]$, \begin{align*} \lvert Tf(t) \rvert &= \left \lvert \int_0^1 K(t,\tau) f(\tau) d\tau \right \rvert \\ &\le \int^1_0 \lvert K(t,\tau) \rvert \lvert f(\tau) \rvert d\tau \\ &\le \left(\int^1_0 \lvert K(t,\tau) \rvert d\tau \right) \| f\|_\infty \\ & \le \left(\int^1_0 M \, d\tau \right)\| f\|_\infty \le M \|f\|_\infty. \end{align*} Since this holds for all $t \in [0,1]$, we can pass to the supremum to see $$\| Tf \|_\infty \le M \|f \|_\infty$$ showing that $T$ is bounded.