Could you help me, please with the following question?
There is a linear functional $A : C_{[0;1]} \rightarrow \mathbb{R}$, such that
$$ Ax=\int_{a}^{b}x(t)\varphi(t)dt $$
where $\varphi$ is a fixed fucntion, $\varphi \in C_{[0;1]}$.
The task is to prove that $ \left \| A \right \| = \int_{a}^{b}\left | \varphi (t) \right |dt $.
The norm in $C_{[0;1]}$ is $||x||=\max_{a \leq t \leq b} |x(t)|$.
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Here is what I did.
First of all we show that the functional is bounded
$\forall x \in C_{[a;b]} ~~~ | Ax | = |\int_{a}^{b}x(t)\varphi(t)dt| \leq \int_{a}^{b} | x(t)| \cdot |\varphi(t)|dt \leq \int_{a}^{b} \max_{a \leq t \leq b} |x(t)| \cdot |\varphi(t)|dt = ||x||\cdot \int_{a}^{b}\left | \varphi (t) \right |dt $
Now i need to prove the inverse inequality or find the sequence of continuous functions with the limit function $x_0$ satisfying $A(x_0)=\int_{a}^{b}\left | \varphi (t) \right |dt~~$?
Or, perhaps, there is some other way? How can i prove that $~~\int_{a}^{b}\left | \varphi (t) \right |dt ~~$ is really a norm?
Thanks in advance.
Best Answer
So you've proven that $\|A\|\leq \|\phi\|_1=\int_a^b|\phi(t)|dt$.
To get the reverse inequality, you have the right idea. You can construct a sequence of continuous functions which approximate the sign function of $\phi$. Note the limit will not be continuous as soon as $\phi$ does not have constant sign. Here is one way to do that.
I assume from now on that $\phi\neq 0$, i.e. $\|\phi\|_\infty>0$, for otherwise everything is trivial. Now set $$ x_n(t):=\frac{\phi(t)}{|\phi(t)|+\frac{1}{n}}. $$ This is clearly continuous and, since $y\longmapsto y/(y+1/n)$ is increasing for $y\geq 0$, we have $$ \|x_n\|_\infty=\frac{\|\phi\|_\infty}{\|\phi\|_\infty+\frac{1}{n}}\longrightarrow 1. $$ On the other hand, we get $$ Ax_n=\int_a^b\frac{\phi(t)^2}{|\phi(t)|+\frac{1}{n}}dt\longrightarrow \int_a^b|\phi(t)|dt $$ since $$ \left|Ax_n-\int_a^b|\phi(t)|dt\right|=\left|\int_a^b \frac{-\phi(t)}{n|\phi(t)|+1}dt\right|\leq \int_a^b \frac{|\phi(t)|}{n|\phi(t)|+1}dt\leq \frac{b-a}{n}. $$ Therefore $$ \|A\|\geq \frac{Ax_n}{\|x_n\|_\infty}\longrightarrow \int_a^b|\phi(t)|dt. $$ This proves the reverse inequality. Hence $\|A\|= \|\phi\|_1=\int_a^b|\phi(t)|dt$.