$ Let~S,~T_n~:(C[0,1],~||~||_โˆž)โ†’(R,~|~|)~be~a~linear~operator.$

$S :=\int_{0}^{1}f(x) dx \\$

$T_n := \frac{1}{n}(\frac{1}{2}f(0) + \sum_{k=1}^{n-1} f(\frac{n}{k}) + \frac{1}{2}f(1))$


I solved S and T_n are bounded, but I have no idea how to solve the latter.


2 Answers 2


We can write $T_nf=\int f d\mu$ where $\mu =\frac 1 {2n} \delta_0+\frac 1 {2n} \delta_1+ \sum\limits_{k=1}^{n-1}\frac 1 n \delta_{n/k}$. Let $m$ be Lebesgue measure on $[0,1]$ and $\nu =m-\mu$. Then $(S-T_n)(f)=\int f d\nu$. By Riesz Representation Theorem the norm of $S-T_n$ is nothing but the total variation of the measure $\mu=m-\mu$. Since $m \perp \mu$ it follows that $\|S-T_n\|=\|m\|+\|\mu\|=1+ \sum\limits_{k=1}^{n-1}\frac 1 n+\frac 1 {2n}+\frac 1 {2n}=2$.

Hints for constructing an elementary proof: $\|T_n-S\|\leq \|T_n\|+\|S\| \le 1+1=2$. Given $\epsilon >0$ construct a piece-wise linear continuous function $f$ of norm $1$ such that $f(x)=-1$ for $x \in\{0,1,\frac 1n, \frac 2 n,..., \frac{n-1} n\}$ and $\int f(x) dx >1-\epsilon$. Then $\|T_n-S\|\geq |Sf-T_nf| >2-\epsilon$.

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    $\begingroup$ @e osa I have added another proof. $\endgroup$ Jul 8, 2020 at 7:13
  • $\begingroup$ Thanks. I totally understand. $\endgroup$ Jul 8, 2020 at 11:35
  • $\begingroup$ Could you tell me why โˆซ๐‘“(๐‘ฅ)๐‘‘๐‘ฅ>1โˆ’๐œ– ? For such a function, โˆซ๐‘“(๐‘ฅ)๐‘‘๐‘ฅ = 1, I think. $\endgroup$ Jul 8, 2020 at 12:17
  • $\begingroup$ You cannot just define $f$ to be $-1$ at a finite number of points and $1$ at all other points. This is becasue $f$ has to be continuous. You have to use an approximation by a continuous function and that is why you cannot get $\int f(x)dx=1$. @eosa $\endgroup$ Jul 8, 2020 at 12:23

I think there are complicated ways to get to a solution, but I want to give you an easy straight-forward one:

First, you should know that the triangle inequality also holds for linear bounded operators. Therefore, we have $$ || S-T_n || \leq ||S|| + ||T_n|| = 1+1=2 .$$

Second, as you surely know, the next step normally involves finding a non-zero function $f$ in $C([0,1],\mathbb{R})$ such that $$ |(S-T_n)f|=2 \cdot ||f||.$$

Unfortunately, this isn't possible.

But there is another way to get the result, namely to find a sequence of non-zero functions $f_m$ such that $$\frac {| (S-T_n)f_m |} {||f_m||} \xrightarrow {k \to \infty} 2.$$

Now you start by considering the case $n=1$. Just start by sketching the functions you'll need. The solution in this case is a sequence of functions that have the values $f(0)=-1=f(1)$ and immediately go to $y=1$ with increasing steepness.

PS: To make it easier, consider functions with norm $1$.


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