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.
 A: You are right, the norm of $T_K$ is indeed
$$\lVert T_K\rVert = \sup_{x\in U_1} \int_{U_2} \lvert K(x,y)\rvert\,dy.$$
However, in general, there is no $f \neq 0$ with $\lVert T_K(f)\rVert = \lVert T_K\rVert\cdot \lVert f\rVert$, so you need to see that you can come arbitrarily close. Given $\varepsilon > 0$, pick an $x\in U_1$ with
$$\int_{U_2} \lvert K(x,y)\rvert\,dy > \lVert T_K\rVert - \varepsilon/2.$$
Then consider the function $h(y) = \overline{\sigma(K(x,y))}$ on $U_2$, where
$$\sigma(z) = \begin{cases} 0 &, z = 0\\ \frac{z}{\lvert z\rvert} &, z \neq 0 \end{cases}$$
is the sign function. We have $\lvert h(y)\rvert \leqslant 1$ for all $y\in U_2$ 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 $\lVert T_K(f)\rVert \geqslant (\lVert T_K\rVert-\varepsilon)\lVert f\rVert$. 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 $\lim\limits_{\delta \to 0} \lVert h_\delta - h\rVert_{L^1(\mathbb{R}^n)} = 0$.
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
$$\left\lvert \int_{U_2} K(x,y)h_\delta(y) - \lvert K(x,y)\rvert\,dy \right\rvert < \varepsilon/2,$$
and thus
$$\lVert T_K(h_\delta)\rVert > (\lVert T_K\rVert - \varepsilon)\lVert h_\delta\rVert.$$

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 $K \in L^\infty (U_1\times U_2)$ instead, so $T_K \colon L^1(U_2) \to L^\infty(U_1)$, we must replace the supremum by an essential supremum.
