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Let $T:C[0,1]\to C[0,1]$ be linear operator defined as follows: $Tf(t)=f(t)+\int_0^t(1-2s)f(s)ds$. Find its norm where norm is is supremum norm.

I am not sure how to do it. It is easy to see that when we take $t\in[0,1/2)$, then we get best result if $f(t)=1$ or $f(t)=-1$. So we can guess than norm is $\leq1+1/4$. But I am not sure how to show that this guess is correct (if it actually is) so I would be thankful for hints how to go on.

Of course if $f$ is always equal to $1$, then we won't get anything better than $1+1/4$. So let's take $\epsilon>0$ and let's check $Tf(t+\epsilon)$ assuming that $f(t+\epsilon)<1$. Then $Tf(t+\epsilon)=f(t+\epsilon)+\int_0^{t+\epsilon}(1-2s)f(s)ds$. We want it to be $<1+1/4$. If $f(t+\epsilon)<0$ it is obvious. Otherwise $f(t+\epsilon)<1$ and $\int_0^{t+\epsilon}(1-2s)f(s)ds<1/4$. So we get what we want. We can use similar argument to show that $|Tf(t)|<1+1/4$ is bounded from both sides. Is it, rather unformal, reasoning on the right track?

SekstusEmpiryk
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  • What norm are you working with? – Piotr Benedysiuk Apr 01 '18 at 13:16
  • @PiotrBenedysiuk, oh I forgot to specify. It is supremum norm. – SekstusEmpiryk Apr 01 '18 at 13:17
  • Quick thought: I get the bound $\left|T\right|_{\infty} \le 3/2$, just from integral inequalities. Then I take $f(s) = -1$ if $s<1/2$ and $f(s) = 1$ if $s\ge 1/2$ to show equality. But this is not continuous, so you need to turn it into a ramp function to get the equality. – user14717 Apr 01 '18 at 13:24
  • I could show it is bounded by norm of $f(t)$ (and some constant) but that about it. I don't have an idea of how to find an expression for the norm. – Piotr Benedysiuk Apr 01 '18 at 13:25
  • @PiotrBenedysiuk, ehh, it is $1-2s$, not $2s-1$. I do not know if it makes much difference. – SekstusEmpiryk Apr 01 '18 at 13:27
  • @user14717, well, I modified expression under integral. But your reasoning is still valid. If $f(s)=1$ on $s<1/2$ and $f(s)=-1$ on $1/2<s<1$, then integral equals to $1/2$. If we set $f(1)=1$, then $Tf(1)=3/2$. Now we just need to work with function sequences and get proper result. Is it ok? – SekstusEmpiryk Apr 01 '18 at 13:37
  • I think that'll work. Here's a list of sigmoid functions: https://en.wikipedia.org/wiki/Sigmoid_function you might be able to integrate one of them analytically (when multiplied by $1-2s$) – user14717 Apr 01 '18 at 13:51

2 Answers2

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Formalizing some ideas by @user14717 and @SekstusEmpiryk from the comments:

\begin{align} \|Tf\|_\infty &\le \|f\|_\infty + \max_{t \in [0,1]} \left|\int_0^t (1-2s)f(s)\,ds\right| \\ &\le \|f\|_\infty + \max_{t \in [0,1]} \int_0^t |1-2s|\underbrace{|f(s)|}_{\le \|f\|_\infty}\,ds \\ &\le \|f\|_\infty + \|f\|_\infty \int_0^1 |1-2s|\,ds \end{align}

We have

$$\int_0^1 |1-2s|\,ds = \int_0^\frac12 (1-2s)\,ds + \int_{\frac12}^1 (2s-1)\,ds = \frac12$$

Therefore $$\|Tf\|_\infty \le \|f\|_\infty \left(1+\frac12\right) = \frac32 \|f\|_\infty$$ so $\|T\| \le \frac32$.

On the other hand, if you define $f_n \in C[0,1]$ as

$$f_n(t) = \begin{cases} 1, &\text{if } t \in \left[0,\frac12 - \frac1n\right]\\ -n\left(t-\frac12\right), &\text{if } t \in \left[\frac12 - \frac1n, \frac12+\frac1n\right]\\ -1, &\text{if } t \in \left[\frac12 - \frac1n, 1-\frac1n\right]\\ 2n\left(x-1+\frac1n\right)-1, &\text{if } t \in \left[1 - \frac1n, 1\right]\\ \end{cases}$$

then we have $\|f_n\|_\infty = 1$ and \begin{align} \|Tf\|_\infty &\ge Tf(1) \\ &= 1 + \underbrace{\int_0^{\frac12-\frac1n} (1-2s)\,ds}_{=\frac14 - \frac1{n^2}} + \underbrace{\int_{\frac12-\frac1n}^{\frac12+\frac1n} (1-2s)f(s)\,ds}_{\ge 0} + \underbrace{\int_{\frac12+\frac1n}^{1-\frac1n} (2s-1)\,ds}_{=\frac14 - \frac1{n}} + \underbrace{\int_{1-\frac1n}^1 (1-2s)f(s)\,ds}_{\ge -\frac1n}\\ &\ge 1+\frac14 - \frac1{n^2} + \frac14 - \frac1{n} -\frac1n\\ &= \frac32 -\frac{2}n - \frac1{n^2} \end{align}

So we conclude

$$\|T\| \ge \frac{\|Tf_n\|_\infty}{\|f_n\|_\infty} \ge \frac32 -\frac{2}n - \frac1{n^2} \xrightarrow{n\to\infty} 1$$

Hence $\|T\| = \frac32$.

mechanodroid
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0

To give a bound note:

$$||T(f)(t)|| \leq ||T(f)(t)||_{\infty} \leq ||f(t)||_{\infty} + || \int\limits_0^t (1-2s)f(s) ds||_{\infty}.$$

The integral we can write out as: $$|| \int\limits_0^t (1-2s)f(s) ds||_{\infty} \leq \int\limits_0^t ||(1-2s)f(s)||_{\infty}ds \leq t ||(1-2t)||_{\infty} ||f(t)||_{\infty} =||f(t)||_{\infty}.$$

So, we can conclude: $||T||\leq 2$

Piotr Benedysiuk
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