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?

  • $\begingroup$ May I inquire as to why you wish to avoid an argument based on Lebesgue integration? It is rather fundamental to operator theory, for example if you were instead considering an integral operator defined on $L^2([0,1])$. $\endgroup$ – Math1000 May 28 '16 at 4:21
  • $\begingroup$ The point is that this problem was presented as an exercise in a mathematical physics course where Lebesgue integration isn't assumed or even mentioned, so I intend to use just what is being really used/assumed in the course. $\endgroup$ – user1620696 May 28 '16 at 18:40

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.

  • $\begingroup$ Wouldn't it be sufficient that $K\in L^\infty([0,1])$? $\endgroup$ – Math1000 May 28 '16 at 4:00
  • 1
    $\begingroup$ I suppose $K \in L^\infty([0,1]\times [0,1])$ should be fine. However, an additional worry is that we need $Tf$ to be continuous whenever $f$ is continuous. When $K \in L^\infty([0,1]\times [0,1])$, we can prove that $Tf$ is continuous using the dominated convergence theorem. If $K \in C([0,1]\times[0,1])$, then we can prove continuity of $Tf$ without using techniques from Lebesgue integration so for the OP's purposes, it may be better to take $K$ continuous. $\endgroup$ – User8128 May 28 '16 at 4:16
  • $\begingroup$ Right, $L^\infty([0,1]^2)$ since that is the domain of $K$. I was asking out of my own curiosity anyway though, thanks! $\endgroup$ – Math1000 May 28 '16 at 4:19
  • $\begingroup$ As I think about this more, I believe we need $t \mapsto K(t,\tau)$ to be a continuous map for a.e. $\tau \in [0,1]$. For any $t \in [0,1]$, we can take a sequence $\{t_n\}$ converging to $t$ and if $Tf$ is to be continuous then we'll need $$\int_0^1 K(t,\tau) f(\tau) d\tau = \int^1_0 \lim_{n \to \infty} K(t_n, \tau) d\tau$$ which means that we need $$K(t, \tau) = \lim_{s \to t} K(s,\tau), \,\,\,\,\, \text{ for a.e. } \tau \in [0,1].$$ $\endgroup$ – User8128 May 28 '16 at 17:14

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