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?

  • $\begingroup$ $dxt$ 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

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

  • $\begingroup$ @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


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?

  • $\begingroup$ 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

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