Prove that linear bounded functional on subspace of C[a,b] that does not attain its norm

Question

Let $L$ be a subspace of space $C[a,b]$ such that $ || f|| = sup_{t\in [a,b]}|f(t)|$. L consists of all $f \in C[a,b]$ such that $ \int_{a}^{\frac{a+b}{2}} f(x)dx=\int_{\frac{a+b}{2}}^{b} f(x)dx $.

a) Prove that L is a vector subspace of $C[a,b] $ and that L is a closed subspace of $C[a,b] $. Then detemine $x^{*} \in (C[a,b])^* $ such that $L = Ker(x^*)$.

b) Prove that $x^*$ from part a) does not obtain its norm on the set {$ f\in C[a,b] , ||f|| \leq 1 $}

I have tried following:

a) I proved that first part of a) and then I tried to prove that L is closed. I took a sequence $\{ f_{n}\}_{n=1}^{\infty} $ , $f_{n} \in L$, such that $f_{n}\rightarrow f $. It means that $sup_{t \in [a,b]}|(f_{n}(t) -f(t) | \rightarrow 0$, $n \rightarrow \infty $, so $\{ f_{n}\}_{n=1}^{\infty} $ uniformly converges to $f$ and $ f \in C[a,b]$. Now we can interchange limit and integral so $ \int_{a}^{\frac{a+b}{2}} f(x)dx = \lim_{n \rightarrow \infty }{ \int_{a}^{\frac{a+b}{2}} f_{n}(x)dx} = \lim_{n \rightarrow \infty }{ \int_{\frac{a+b}{2}}^{b} f_{n}(x)dx} =\int_{\frac{a+b}{2}}^{b} f(x)dx $ and therefore, $f \in L$. We can also conclude that L is complete space because it is closed.

In the second part of a), can we simply put that $x^* (f)= \int_{a}^{\frac{a+b}{2}} f(x)dx - \int_{\frac{a+b}{2}}^{b} f(x)dx $?

b) I didn't really know how to prove this, so can someone check what I did so far and help solving the part b)? Thanks in advance.

Answer 1

Your answer to $(a)$ all looks fine.

For (b), you should start by computing $\|x^*\|$. The trivial upper bound is given by $$|x^*(f)| \leq \bigg|\int_{a}^{\frac{a+b}{2}} f(x)dx\bigg| + \bigg| \int_{\frac{a+b}{2}}^{b} f(x)dx\bigg| \leq \int_a^b |f(x)| dx \leq (b-a) \|f\|_\infty$$ so that $\|x^*\| \leq (b-a)$.

Next, show that in fact $\|x^*\| = (b-a)$. This means that you want to find a sequence $f_n$ with $\|f_n\|_\infty = 1$ such that $x^*(f_n) \to b-a$. If we forget the continuity assumption, it's clear that to get $x^*(f) = b-a$, we'd want to put $f = 1_{[a, \frac{a+b}{2}]} - 1_{(\frac{a+b}{2}, b]}$. Since this $f$ is discontinuous we can't do this but we can approximate this $f$ by continuous functions. Set $$f_n(x) = \begin{cases} 1, \qquad & x \in [a, \frac{a+b}{2} - \frac1n] \\ - 1, \qquad & x \in (\frac{a+b}{2} + \frac1n, b], \\ 1 - n (x - \frac{a+b}{2}), \qquad & \text{ otherwise } \end{cases}$$ Then $\|f_n\|_\infty = 1$ and a simple calculation shows that $|x^*(f_n)| = |b-a - \frac3n| \to b-a$ as $n \to \infty$ so $\|x^*\| = b-a$.

It only remains to show that there is no $f \in C([a,b])$ with $\|f\|_\infty = 1$ such that $x^*(f) = b-a$. I already mentioned the intuition for this earlier. An $f$ with that property would satisfy $$\int_a^{\frac{a+b}{2}} f(x) dx = \frac{b-a}{2} = - \int_{\frac{a+b}{2}}^b f(x) dx$$ since e.g. $$\bigg |\int_a^{\frac{a+b}{2}} f(x) dx \bigg | \leq \frac{b-a}{2} \|f\|_\infty = \frac{b-a}{2}.$$

This means that $f(x) = 1$ on $[a, \frac{a+b}{2})$ and $-1$ on $(\frac{a+b}{2}, b]$ and hence is discontinuous which is a contradiciton.

Indeed, if $f$ is continuous, $\|f\|_\infty \leq 1$ and $f$ does not take those values then without loss of generality, by continuity of $f$ there is $x \in [a, \frac{a+b}{2})$ and $\varepsilon > 0$ such that $(x-\varepsilon, x + \varepsilon) \subseteq [a, \frac{a+b}{2})$ and $y \in (x-\varepsilon, x + \varepsilon)$ implies that $f(y) < 1-\varepsilon$. Then since $f(x) \leq 1$, it is easy to bound $$\int_a^{\frac{a+b}{2}} f(x) dx \leq \int_{[a, \frac{a+b}{2})\setminus (x-\varepsilon, x + \varepsilon)} f(x) dx + \int_{(x-\varepsilon, x + \varepsilon)}f(x) dx \leq \frac{b-a}{2} - 2 \varepsilon + 2\varepsilon (1-\varepsilon) < \frac{b-a}{2}$$ which is a contradiction.

Answer 2

Your part a) solution is correct. For b) you need firstly prove that $\|f\|=b-a$. Now suppose that there exists $f_0$ such that $\|f_0\|\leq1$ and $x^\ast(f_0)=b-a$. Since $-1\leq f_0(x)\leq1$, then $\left|\displaystyle\int\limits_{a}^{\frac{a+b}{2}} f_0(x)dx\right|\leq\dfrac{b-a}{2}$, and $\left|\displaystyle\int\limits_{\frac{a+b}{2}}^{b} f_0(x)dx\right|\leq\dfrac{b-a}{2}$. So $x^\ast(f_0)=b-a$ only if $\displaystyle\int\limits_{a}^{\frac{a+b}{2}} f_0(x)dx=\dfrac{b-a}{2}$, and $\displaystyle\int\limits_{\frac{a+b}{2}}^{b} f_0(x)dx=-\frac{b-a}{2}$.

Notice, that if exists point $x_0\in\left[a,\frac{a+b}{2}\right]$ such that $f(x_0)<1$, then, by continuous of $f_0(x)$, there is a neighborhood of $x_0$ at which $f_0(x)<1$, thus $\displaystyle\int\limits_{a}^{\frac{a+b}{2}} f_0(x)dx<\dfrac{b-a}{2}$. Therefore $\displaystyle\int\limits_{a}^{\frac{a+b}{2}} f_0(x)dx=\dfrac{b-a}{2}$ only if $f_0(x)\equiv1$ on $\left[a,\frac{a+b}{2}\right]$. Similarly $\displaystyle\int\limits_{\frac{a+b}{2}}^{b} f_0(x)dx=-\frac{b-a}{2}$ only if $f_0(x)\equiv-1$ on $\left[\frac{a+b}{2},b\right]$. So, $f_0(x)$ is discontinuous function at $x=\frac{a+b}{2}$, which is contradiction.