Exercise 1: The operator norm
Let $L(X, Y)$ be the space of bounded linear operators between two normed spaces $X \neq{0}$ and $Y$, equipped with the operator norm $|T|=\sup {\substack{x \in X \ x \neq 0}} \frac{|T x|}{|x|}$. (a) Prove the equalities $$|T|=\inf {C \geq 0:|T x| \leq C|x| \text { for all } x \in X}=\sup {\substack{x \in X \|x|=1}}|T x|=\sup _{\substack{x \in X \|x| \leq 1}}|T x| .$$
(b) Prove that $L(X, Y)$ is a vector space and prove that $|\cdot|$ is a norm on $L(X, Y)$.

Solution:
(a) Note that the estimate
$$\sup {\substack{x \in X \|x|=1}}|T x| \leq \sup {\substack{x \in X \|x| \leq 1}}|T x|$$
is clear, since we are taking suprema of the same quantity over a larger set. Now let $x \in X$ with $|x| \leq 1$. Then if $x=0$, we have $|T x|=0 \leq|T|$. Otherwise,
$$|T x| \leq \frac{|T x|}{|x|} \leq|T|$$
Thus, taking a supremum over all $x \in X$ with $|x| \leq 1$, we conclude that sup $\underset{|x| \leq 1}{ }|T x| \leq$ $|T|$.
It remains to show that
$$|T| \leq \inf {C \geq 0:|T x| \leq C|x| \text { for all } x \in X} \leq \sup _{\substack{x \in X \|x|=1}}|T x|$$
to conclude the equalities.
For the first inequality, suppose that $C \geq 0$ satisfies the property that $|T x| \leq C|x|$ for all $x \in X$. Then for all $x \in X \backslash{0}$ we have
$$\frac{|T x|}{|x|} \leq C$$
Taking a supremum over all such $x$ yields $|T| \leq C$. Then, taking an infimum over all such $C \geq 0$ proves the first inequality.

For the second inequality, note that if $y \in X \backslash{0}$, then
$$|T y|=\left|T\left(\frac{y}{|y|}\right)\right||y| \leq \sup {\substack{x \in X \|x|=1}}|T x||y| .$$ Since the inequality $|T y| \leq \sup {\substack{x \in X \|x|=1}}|T x||y|$ also holds for $y=0$, we conclude that
$$\inf {C \geq 0:|T x| \leq C|x| \text { for all } x \in X} \leq \sup _{\substack{x \in X \|x|=1}}|T x|,$$
as desired. The assertion follows.
(b) We only prove that if $T, S \in L(X, Y)$, then also $T+S \in L(X, Y)$ with $|T+S| \leq|T|+|S|$.
Let $x \in X$. Then
$$|(T+S) x|=|T x+S x| \leq|T x|+|S x| \leq|T||x|+|S||x|=(|T|+|S|)|x|$$
This proves that $T+S \in L(X, Y)$. Since, by the second characterization of part (a), the norm $|T+S|$ is given by the smallest constant $C \geq 0$ such that $|(T+S) x| \leq C|x|$ for all $x \in X$, it follows from (1) that $|T+S| \leq|T|+|S|$. The result follows.

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Exercise 3: Integral operators
Let $k$ be a continuous function $k:[0,1] \times[0,1] \rightarrow \mathbb{R}$. For each $f \in C([0,1]$ we define a function $T f:[0,1] \rightarrow \mathbb{R}$ by
$$T f(t):=\int_0^1 k(t, s) f(s) \mathrm{d} s$$
(a) Show that $T f \in C([0,1])$ for $f \in C([0,1])$ and, moreover, show that $T: C([0,1]) \rightarrow C([0,1])$ is a linear operator.
(b) Show that
$$|T|=\sup _{t \in[0,1]} \int_0^1|k(t, s)| \mathrm{d} s$$

Solution:
(a) Let $f \in C([0,1])$. We will show that $T f$ is in fact a uniformly continuous function. We set $|f|_1:=\int_0^1|f(s)| \mathrm{d} s$. Note that this constant is finite, since
$$|f|_1=\int_0^1|f(s)| \mathrm{d} s \leq|f|_{\infty} \int_0^1 \mathrm{~d} s=|f|_{\infty}<\infty$$ Next, let $\varepsilon>0$. Since $k$ is a continuous function on the compact set $[0,1] \times[0,1]$, it is actually uniformly continuous. Thus, we can choose $\delta>0$ such that whenever $\left|(t, s)-\left(t^{\prime}, s^{\prime}\right)\right|<\delta$, we have $\left|k(t, s)-k\left(t^{\prime}, s^{\prime}\right)\right|<\frac{\varepsilon}{1+|f|_1}$.

If $t, t^{\prime} \in[0,1]$ satisfy $\left|t-t^{\prime}\right|<\delta$, then for each $s \in[0,1]$ we have $\left|(t, s)-\left(t^{\prime}, s\right)\right|=\left|\left(t-t^{\prime}, 0\right)\right|=$ $\left|t-t^{\prime}\right|<\delta$. Hence,
\begin{aligned} \left|T f(t)-T f\left(t^{\prime}\right)\right| & =\left|\int_0^1\left(k(t, s)-k\left(t^{\prime}, s\right)\right) f(s) \mathrm{d} s\right| \leq \int_0^1\left|k(t, s)-k\left(t^{\prime}, s\right) | f(s)\right| \mathrm{d} s \ & \leq \frac{\varepsilon}{1+|f|_1} \int_0^1|f(s)| \mathrm{d} s=\frac{\varepsilon|f|_1}{1+|f|_1}<\varepsilon . \end{aligned}
Thus, $T f$ is uniformly continuous. In particular $T f \in C([0,1])$.
We point out here that the only property we needed of $f$ is that it is integrable over $[0,1]$ to conclude the continuity of $T f$.
The linearity of $T$ follows from the linearity of the integral.
(b) Set $A:=\sup {t \in[0,1]} \int_0^1|k(t, s)|$ ds. We will show that $|T| \leq A$ and $A \leq|T|$ to conclude that $|T|=A$. For the first inequality, note that for all $t \in[0,1]$ we have $$|T f(t)| \leq \int_0^1|k(t, s)||f(s)| \mathrm{d} s \leq \int_0^1|k(t, s)| \mathrm{d} s|f|{\infty} \leq \sup {t \in[0,1]} \int_0^1|k(t, s)| \mathrm{d} s|f|{\infty}=A|f|_{\infty}$$

Taking a supremum over all $t \in[0,1]$ yields $|T f|_{\infty} \leq A|f|_{\infty}$. Thus, $|T| \leq A$, as desired.
For the converse inequality, note that since $k$ is continuous, so is $|k|$. Hence, by the same argument as in part (a), the map $t \mapsto \int_0^1|k(t, s)| \mathrm{d} s$ for $t \in[0,1]$ is also continuous. Since $[0,1]$ is compact, this means that there is a $t_0 \in[0,1]$ where the supremum of this function is attained, i.e., $A=\sup {t \in[0,1]} \int_0^1|k(t, s)| \mathrm{d} s=\int_0^1\left|k\left(t_0, s\right)\right| \mathrm{d} s$. Now for $n \in \mathbb{N}$ we define $$f_n:[0,1] \rightarrow \mathbb{R}, \quad f_n(s):=\frac{k\left(t_0, s\right)}{\left|k\left(t_0, s\right)\right|+\frac{1}{n}} .$$ Note that then $\left|f_n\right|{\infty} \leq 1$ so that
$$\left|T f_n\left(t_0\right)\right| \leq\left|T f_n\right|_{\infty} \leq\left|f_n\right|_{\infty}|T| \leq|T|$$
But
\begin{aligned} \left|T f_n\left(t_0\right)\right| & =\int_0^1 \frac{\left|k\left(t_0, s\right)\right|^2}{\left|k\left(t_0, s\right)\right|+\frac{1}{n}} \mathrm{~d} s=\int_0^1\left|k\left(t_0, s\right)\right|\left(1-\frac{\frac{1}{n}}{\left|k\left(t_0, s\right)\right|+\frac{1}{n}}\right) \mathrm{d} s \ & =\int_0^1\left|k\left(t_0, s\right)\right| \mathrm{d} s-\frac{1}{n} \int_0^1 \frac{\left|k\left(t_0, s\right)\right|}{\left|k\left(t_0, s\right)\right|+\frac{1}{n}} \mathrm{~d} s \rightarrow \int_0^1\left|k\left(t_0, s\right)\right| \mathrm{d} s=A \end{aligned}
as $n \rightarrow \infty$. Thus, letting $n \rightarrow \infty$ in (3), we conclude that $A \leq|T|$. Thus, we have $|T|=A$, proving the result.

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