Editing Unbounded operator (section)

== Example ==
Let {{math|''C''([0, 1])}} denote the space of continuous functions on the unit interval, and let {{math|''C''<sup>1</sup>([0, 1])}} denote the space of continuously differentiable functions. We equip <math>C([0,1])</math> with the supremum norm, <math>\|\cdot\|_{\infty}</math>, making it a Banach space. Define the classical differentiation operator {{math|{{sfrac|''d''|''dx''}} : ''C''<sup>1</sup>([0, 1]) → ''C''([0, 1])}} by the usual formula:

: <math> \left (\frac{d}{dx}f \right )(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}, \qquad \forall x \in [0, 1].</math>

Every differentiable function is continuous, so {{math|''C''<sup>1</sup>([0, 1]) ⊆ ''C''([0, 1])}}. We claim that {{math|{{sfrac|''d''|''dx''}} : ''C''([0, 1]) → ''C''([0, 1])}} is a well-defined unbounded operator, with domain {{math|''C''<sup>1</sup>([0, 1])}}. For this, we need to show that <math>\frac{d}{dx}</math> is linear and then, for example, exhibit some <math>\{f_n\}_n \subset C^1([0,1])</math> such that <math>\|f_n\|_\infty=1</math> and <math>\sup_n \|\frac{d}{dx} f_n\|_\infty=+\infty</math>.

This is a linear operator, since a linear combination {{math|''a&thinsp;f&thinsp;'' + ''bg''}} of two continuously differentiable functions {{math|&thinsp;''f''&thinsp;, ''g''}} is also continuously differentiable, and

:<math>\left (\tfrac{d}{dx} \right )(af+bg)= a \left (\tfrac{d}{dx}  f \right ) + b \left (\tfrac{d}{dx} g \right ).</math>

The operator is not bounded. For example,

:<math>\begin{cases} f_n : [0, 1] \to [-1, 1] \\ f_n(x) = \sin (2\pi n x) \end{cases}</math>

satisfy

:<math> \left \|f_n \right \|_{\infty} = 1,</math>

but

:<math> \left \| \left (\tfrac{d}{dx}  f_n \right ) \right \|_{\infty} = 2\pi n \to \infty</math>
as <math>n\to\infty</math>.

The operator is densely defined (which can be shown by the Weierstrass approximation theorem, since the set of polynomial functions on [0,1] is contained in {{math|''C''<sup>1</sup>([0, 1])}}, while also being dense in {{math|''C''([0, 1])}}) and closed.

The same operator can be treated as an operator {{math|''Z'' → ''Z''}} for many choices of Banach space {{mvar|Z}} and not be bounded between any of them. At the same time, it can be bounded as an operator {{math|''X'' → ''Y''}} for other pairs of Banach spaces {{math|''X'', ''Y''}}, and also as operator {{math|''Z'' → ''Z''}} for some topological vector spaces {{mvar|Z}}.{{clarify|reason=Why the shift from Banach spaces to topological vector spaces? What is a bounded operator between topological vector spaces?|date=May 2015}} As an example let {{math|''I'' ⊂ '''R'''}} be an open interval and consider

:<math>\frac{d}{dx} : \left (C^1 (I), \|\cdot \|_{C^1} \right ) \to \left ( C (I), \| \cdot \|_{\infty} \right),</math>

where:

:<math>\| f \|_{C^1} = \| f \|_{\infty} + \| f' \|_{\infty}.</math>