Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $ E = \mathcal B([0,1], \Bbb R) $ with supremum norm.

Now I can define function $ F:E \ni f \rightarrow ||f||^2 - f(0) \in \Bbb R$

My task:
1)Show the differentiability of $F$ in: $ f_0: [0,1] \ni x \rightarrow [x] \in \Bbb R $;
2)Calculate the derivative in $f_0$ and it's norm;
3)Check, if subspace of $E$ : $ H = \{ f \in E | f(0) = f(1) \} $ is a Banach space.

This is what I've done already:

$ \lim _{||h|| \rightarrow 0 } \frac{| F (f_{0}+h) - F(f_0) - L(h)|}{||h||} =\lim _{||h|| \rightarrow 0 } \frac{| \ ||f_0+h||^2 - (f_0+h)(0) - ||f_0||^2 + f_0(0)- L(h)|}{||h||} = \lim _{||h|| \rightarrow 0 } \frac{| \ [1+ h(1)]^2 - h(0) - 1 - L(h)|}{||h||} = \lim _{||h|| \rightarrow 0 } \frac{| \ 1+ 2h(1) + [h(1)]^2 - h(0) - 1 - L(h)|}{||h||} = \lim _{||h|| \rightarrow 0 } \frac{| \ 2h(1) + [h(1)]^2 - h(0) - L(h)|}{||h||} = 0$
(I think) $L(h) = 2h(1)-h(0)$, because $[h(0)]^2$ is too small ( something with $o(h)$ what I do not understand :( )

How to calculate the norm? I have no idea a.t.m.

About $H$: if I make a Cauchy sequence of functions ($n \in \Bbb N_+$):
$f_n(x) = \begin{cases}\frac{2^{n-1}}{2^n -1}x \text{, if } x\in[0, \frac{2^n -1}{2^n}] \\ 1-\frac{2^{n-1}}{2^n -1}x \text{, if } x \in (\frac{2^n -1}{2^n},1]\end{cases}$
This converges to a function:
$f(x)= \frac{1}{2}x$, where $f(0) = 0$ and $f(1)= \frac{1}{2}$
Is this done right?

share|cite|improve this question

(1) Your derivative $L$ is correct. But you should add why you know that $\|f_0+h\| = 1+h(1)$. This is actually the better place to start:

If $\|h\|<1/2$, then $|f_0+h|<1/2$ on $[0,1)$ and $|f_0+h|>1/2$ at $1$. Hence, $\|f_0+h\|=f_0(1)+h(1)=1+h(1)$.

Then you see that $F(f_0+h) = 1+2h(1)+h(1)^2 -h(0) $ for all such $h$. The derivative is a linear approximation to $F(f_0+h)-F(f_0)$. Since $$F(f_0+h)-F(f_0)=2h(1)+h(1)^2 -h(0)$$ we take $L(h) = 2h(1) -h(0)$. The term $h(1)^2$, when divided by $\|h\|$, will have limit $0$ as $h\to 0$. This is precisely what the notation $h(1)^2 = o(h)$ means.

(2) The norm of linear function $L$ is $3$. On one hand, $$|L(h)|\le 2|h(1)|+|h(0)| \le 3\|h\| \tag{1}$$ On the other, there exist functions $h$ for which (1) turns into equality. Namely, $h(x)=2x-1$; or any function such that $h(1)=1$, $h(0)=-1$, and $\|h\|=1$.

(3) The sequence you have does not belong to $H$, because $f_n(1)\ne f_n(0)$. Moreover, no such counterexamples exist: the space $H$ is closed in $E$. To see this, note that $f\mapsto f(0)-f(1) $ is a bounded linear functional on $E$ (its norm is $2$). The kernel of a bounded linear functional is a closed subspace.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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