§ Differentiation

• Simplest example is differentiation.
• Let $C^0[0, 1]$ be continuous functions on interval, and $C^1[0, 1]$ be differentiable functions on the interval.
• We equip both spaces with sup norm / infty norm.
• Consider the differentiation operator $\partial : C^1[0, 1] \to C[0, 1]$.
• Since every differentiable function is continuous, we have that $C^[0, 1] \subseteq C[0, 1]$
• Clearly differentiation is linear (well known).
• To see that the operator is not bounded, consider the sequence of functions $f_n(x) \equiv sin(2\pi nx)$.
• We have that $||f_n||_\infty = 1$ for ann $n$, while the $||\partial_x f_n||_\infty \to \infty$, so clearly, there is no constant $M$ such that $||\partial f_n(x)|| \leq M ||f_n(x)||$. Thus, the operator is unbounded.
• Note that in this definition, the space $C^1[0, 1]$ is not closed, as there are sequences of differentiable functions that coverge to non differentiable functions. Proof: polynomials which are differentiable functoins are dense in the full space of continuous functions.
• Thus, in the case of an unbounded operator, we consider $L : U \to X$ where $U$ is some subspace of $X$, not ncessarily closed!
• If we ask for an everywhere defined operator, then constructing such operators $L : X \to X$ needs choice.

§ Nonconstructive example

• Regard $\mathbb R$ as a normed vector space over $\mathbb Q$. [Cannot call this a banach space, since a banach space needs base field $\mathbb R$]
• Find an algebraic basis $B$ containing the numbers $1$ and $\pi$ and whatever else we need.
• define a function $f: \mathbb R \to \mathbb R$ such that $f(\pi) = 0$, and $f(1) = 1$, and extend everywhere else by linearity.
• Now let $p_i \in \mathbb Q$ be a sequence of rationals that converge to $\pi$. Then $f(p_i) = 0$, and thus $\lim_i f(p_i) = 0$, while $f(\lim_i p_i) = f(\pi) = 1$. This shows that $f$ is not continuous, but is linear.