# 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.