Norm of an integral operator on the space of continuous functions

The following is a question I came up with when I was studying the same problem in dimension 1 (for which also I have the questions that follows) but I put in generality.

Let $U_1, U_2 subset mathbb R^n$ bounded, $displaystyle f:U_2 to
mathbb R$, and $displaystyle K: overline U_1 times overline U_2
to mathbb R$ continuous. For $displaystyle x in U_1 $ define the
integral operator $displaystyle T_K : C(U_2) to C(U_1) $ defined by
$$ (T_K f)(x):= int_{U_2} K(x,y) f(y) text{d} y $$

I think that it is $ displaystyle |T_K | = sup_{x in U_1} int_{U_2} |K(x,y)| text{d} y $, but I am don’t know how to prove it. I have shown that this $sup$ is the upper bound for the norm of the operator, but I can’t find an $f$ that achieves it.

Also I have one more question:

If instead of $f$ being contninuous we suppose that $f$ is only
Lebesgue measurable, what is changing, or everything is the same.

Thank you.


You are right, the norm of $T_K$ is indeed

$$lVert T_KrVert = sup_{xin U_1} int_{U_2} lvert K(x,y)rvert,dy.$$

However, in general, there is no $f neq 0$ with $lVert T_K(f)rVert = lVert T_KrVertcdot lVert frVert$, so you need to see that you can come arbitrarily close. Given $varepsilon > 0$, pick an $xin U_1$ with

$$int_{U_2} lvert K(x,y)rvert,dy > lVert T_KrVert – varepsilon/2.$$

Then consider the function $h(y) = overline{sigma(K(x,y))}$ on $U_2$, where

$$sigma(z) = begin{cases} 0 &, z = 0\ frac{z}{lvert zrvert} &, z neq 0 end{cases}$$

is the sign function. We have $lvert h(y)rvert leqslant 1$ for all $yin U_2$ and

$$int_{U_2} K(x,y)h(y),dy = int_{U_2} lvert K(x,y)rvert,dy.$$

If $h$ happens to be continuous, we already have found a function in $C(U_2)$ with $lVert T_K(f)rVert geqslant (lVert T_KrVert-varepsilon)lVert frVert$. But generally, $h$ will not be continuous, and thus we need to approximate it by continuous functions. Extend $h$ to a function on all of $mathbb{R}^n$ by the value $0$ outside $U_2$. Then choose a mollifier function $eta$ with $0 leqslant eta(t)$, $int_{mathbb{R}^n}eta(t),dt = 1$, and $eta$ continuous. For $delta > 0$ define

$$h_delta(y) = int_{mathbb{R}^n} eta(t)h(y- delta t),dt.$$

The functions $h_delta$ are continuous, satisfy $lvert h_delta(y)rvert leqslant 1$ for all $y$, and $limlimits_{delta to 0} lVert h_delta – hrVert_{L^1(mathbb{R}^n)} = 0$.

From the latter, it follows that

$$int_{U_2} K(x,y) h_delta(y),dy xrightarrow{delta to 0} int_{U_2} K(x,y) h(y),dy$$

since $K$ is bounded. Thus, for small enough $delta > 0$, we have

$$leftlvert int_{U_2} K(x,y)h_delta(y) – lvert K(x,y)rvert,dy rightrvert < varepsilon/2,$$

and thus

$$lVert T_K(h_delta)rVert > (lVert T_KrVert – varepsilon)lVert h_deltarVert.$$

If instead of $f$ being contninuous we suppose that $f$ is only Lebesgue measurable, what is changing, or everything is the same.

We need to demand that $f$ is integrable, not only measurable. The result remains the same, but in that case, we can directly work with $h$ and don’t need to approximate it by continuous functions.

If we drop the requirement that $K$ is continuous, and consider $K in L^infty (U_1times U_2)$ instead, so $T_K colon L^1(U_2) to L^infty(U_1)$, we must replace the supremum by an essential supremum.

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