2
$\begingroup$

I found wierd (for me) example about computing norm of the functional. We have $$\varphi : \mathcal{C}([0,1])\ni f\mapsto f(1)\in\mathbb{R}$$ with the norm in $\mathcal{C}([0,1])$ given by: $$\Vert f\Vert=\sup_{x\in[0,1]}|f(x)| +\int_0^1 |f(x)|dx.$$

Clearly, it is well-defined and linear so we find a constance of the continuity: $$\Vert\varphi(f)\Vert=|f(1)|\le\sup_{x\in[0,1]}|f(x)|\le\sup_{x\in[0,1]}|f(x)| +\int_0^1 |f(x)|dx=\Vert f\Vert.$$

So $\Vert\varphi(f)\Vert\le M\Vert f\Vert$ for $M=1$. I tried to estimate it in the other way but I failed.

And now we want to find such $f$ that attains the norm, i.e. $$\Vert f\Vert=\sup_{x\in[0,1]}|f(x)| +\int_0^1 |f(x)|dx=1$$ and $$ \Vert\varphi(f)\Vert=M=1.$$

And now I have any idea what to do, cause we have such inequality: $$\Vert\varphi(f)\Vert=|f(1)|\le\sup_{x\in [0,1]}|f(x)|$$ clearly, and since $\int_0^1 |f(x)|dx$ is grater than or equal to $0$ (but $0$ on the right hand side would be attain only for $f\equiv 0$, which contradicts with $\Vert f\Vert=1$), we get a strict inequality: $$1=\Vert\varphi(f)\Vert=|f(1)|<\sup_{x\in [0,1]}|f(x)|+\int_0^1 |f(x)|dx=\Vert f\Vert=1$$

So how to finish it? Some tricky solution or my thinking is wrong completely?

$\endgroup$
  • $\begingroup$ $dxt$ should it be $dx$ or what is $t$? $\endgroup$ – mathreadler Feb 25 '17 at 10:16
  • $\begingroup$ it was mistake obviously, it should be $dx$ ;) $\endgroup$ – Yelon Feb 25 '17 at 10:18
2
$\begingroup$

If you consider, for all $n\ge1$ the function :

$$f_n:[0,1]\to\mathbb{R},x\mapsto\cases{0\quad\mathrm{if}\,0\le x\le 1-\frac 1n\cr n\left(x-1+\frac 1n\right)\quad\mathrm{otherwise}}$$

you can see that $\varphi(f_n)=1$ and $\Vert f_n\Vert=1+\frac 1{2n}$

Hence $$\lim_{n\to\infty}\frac{\vert\varphi(f_n)\vert}{\Vert f_n\Vert}=1$$

$\endgroup$
  • $\begingroup$ @mathcounterexamples: same idea at the same time :) $\endgroup$ – Adren Feb 25 '17 at 10:34
  • 2
    $\begingroup$ Indeed! In fact, an even easier sequence of functions to consider would be $f_n(x)=x^n$. $\endgroup$ – mathcounterexamples.net Feb 25 '17 at 10:40
2
$\begingroup$

Hint

Consider the sequence of functions $(f_n)$ defined by $$f_n(x)=\begin{cases} 0 & \text{ for } x \in [0,1-\frac{1}{n}]\\ nx-n+1 & \text{ for } x \in [1-\frac{1}{n},1] \end{cases}$$

For all $n \in \mathbb N$, you have $f_n \in \mathcal{C}([0,1])$ and $f_n(1)=1$.

What is the value of $\Vert f_n \Vert$? What can you conclude?

Additional remark: There is no function $g \in \mathcal{C}([0,1])$ such that $\Vert g \Vert = 1$ and $\varphi(g)=1$. In other terms, the supremum of $\varphi$ is not attained on the unit closed ball. Do you see why?

$\endgroup$
  • $\begingroup$ about remark: does not the last inequality in my post show this contradiction? $\endgroup$ – Yelon Feb 25 '17 at 12:07
  • $\begingroup$ @Yelon Yes, that is true. $\endgroup$ – mathcounterexamples.net Feb 25 '17 at 12:35

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.