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So, I was practicing some problems and considered the space $X = C[a,b]$ with the $L_{1}$-norm. I consider the operator

$$Tf(x) = \int_a^b k(x,y)f(y)\,dy$$, where $k(x,y)$ is continuous in both of its variables.

So, I find this operator is bounded:

$$\|T\| \le \operatorname{max}_{a\le x \le b} \int_a^b |k(x,y)|\,dy$$

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Since $k$ is continuous on a compact domain, it reaches its maximum at some $x_{0}$ i.e there is some $x_{0}$ such that $$\operatorname{max}_{a\le x \le b} \int_a^b |k(x,y)|\,dy = \int_a^b |k(x_{0},y)|\,dy$$.

In analogy with the finite dimensional matrix case, where the proof for the 1-norm involves just the vector in that direction, I want to approximate the delta function with a continuous function, say $f_{n} = \frac{n}{\pi(1+n^2(x-x_{0})^2)}$. However, this is where I get confused. How would I estimate this as I want to get

$$\|T\| \ge \operatorname{max}_{a\le x \le b} \int_a^b |k(x,y)|\,dy$$. Or am I just doing this the entirely wrong way?

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You are doing it the right way, but there are easier choices of $f_n$.

Let $\phi(t) = \max(1-|t|,0)$, and $\phi_n(t) = n \phi(n t)$. Note that $\phi_n \ge 0$, $\int \phi_n = 1$, and $\operatorname{supp} \phi_n = [-\frac{1}{n}, \frac{1}{n}]$.

Let $\overline{K} = \sup_t \int |k(s,t)| ds$, $\epsilon>0$ and suppose $t_0$ is such that $\int |k(s,t_0)| ds > \overline{K}-\frac{\epsilon}{2}$. By continuity, we may assume that $t_o \in (a,b)$ (that is, excluding $a,b$, purely for convenience).

$k$ is continuous, and $[a,b]^2$ is compact, hence $k$ is uniformly continuous. Choose $\delta>0$ such that if $|t-t_0| < \delta$, then $|k(s,t)-k(s,t_0)| < \frac{\epsilon}{2(b-a)}$, for all $s$. Now let $f_n(t) = \phi_n(t-t_0)$, and choose $n$ large enough so that $\operatorname{supp} f_n \subset B(t_0,\delta)$. (Note that $\|f_n\|_1 = 1$.)

Then we have

\begin{eqnarray}

\left| \int (k(s,t)-k(s,t_0)) f_n(t) dt \right| &\le& \int |k(s,t)-k(s,t_0)| f_n(t) dt \\

&=& \int_{t_0-\delta}^{t_0+\delta} |k(s,t)-k(s,t_0)| f_n(t) dt \\

&\le& \frac{\epsilon}{2(b-a)} \int_{t_0-\delta}^{t_0+\delta} f_n(t) dt \\

&=& \frac{\epsilon}{2(b-a)}

\end{eqnarray}

Noting that $\int k(s,t_0) f_n(t) dt = k(s,t_0)$, we have

$|\int k(s,t) f_n(t) dt| \ge |\int k(s,t_0) f_n(t) dt|-\frac{\epsilon}{2(b-a)} = |k(s,t_0)|-\frac{\epsilon}{2(b-a)} $.

Finally, $\int|Tf_n| = \int |\int k(s,t) f_n(t) dt| ds \ge \int |k(s,t_0)| ds- \frac{\epsilon}{2} \ge \overline{K} – \epsilon$.

Since $\epsilon>0$ was arbitrary, we have $\|T\| \ge \overline{K}$.

Since you already know that $\|T\| \le \overline{K}$, you have the desired result.

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