2
$\begingroup$

Let $C^{(1)}[0,1]$ be the space of all continuous functions on $[0,1]$ that have continuous derivative. (This space is called the space of continuously differentiable functions.) Define a norm $\|\cdot \|$ on $C^{(1)}[0,1]$ by $\| f \|=\max _{t \in T}|f(t)|.$ Show that the map $$B(f) = f^{\prime}(0)$$ defines a linear functional on the normed vector space $(C^{(1)}[0,1], \| \cdot \|)$ that is not bounded.

My attempt:

For any $f,g \in C^{(1)}[0,1],$ we have $B(f+g) = (f + g)^{\prime}(0) = f^{\prime}(0) + g^{\prime}(0) = B(f) + B(g).$ Therefore, $B$ is a linear functional.

To show that $B$ is not bounded, we wish to show that for any $M > 0$, there exists $f \in C^{(1)}[0,1]$ such that $\| f \| \leq 1$ but $|B(f)| > M.$

Let $M > 0 $ be given. I have trouble constructing a function with maximum $\leq1$ and has derivative at $x = 0$ greater than $M.$

I plan to construct $f$ which looks similar to $\sin(x)$. For any given $M > 0$, I can draw $\sin(x)$ such that it is above the line $y = Mx$. However, I do not know how to formulate this phenomena using mathematical symbols.

$\endgroup$
2
$\begingroup$

Define $ f_n(x) = \sin (nx) $ then $\sup|f_n(x)| =1$ however $f_n'(x) = n\cos nx$ hence $$|f'_n(0)| =n.$$

$\endgroup$
  • $\begingroup$ Nice. This is exactly what I want. $\endgroup$ – Idonknow May 24 '17 at 2:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.