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Take $X=C([0,1])$ with the uniform norm, $\|f\|=\sup_{x\in[0,1]}|f(x)|$, and define the operator $T:X\to X$ by, $$T(f)(x)=f(x)-\int_0^1f(s)ds$$ Find $\|T\|$.

I was hoping to solve this problem using directly the definition of the operator norm, namely,

$$\|T\|=\sup\{\|Tx\|:x\in X,\|x\|\le1\}=\{\frac{\|Tx\|}{\|x\|}:x\in X, x\neq0\}$$

However I am not sure how to begin. I have already shown that $T$ in this case is a bounded linear operator, being bounded above by $k=2$, in the sense that,

$$\|Tf\|\le2\|f\|,\,\,\,\forall f\in X$$

The solution we done in class involved constructing a sequence of functions $(f_n)_n^\infty\subset X$ such that $Tf_n\to2$. Is there any way to find $\|T\|$ without constructing such a sequence, by using the definition of the operator norm?

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  • $\begingroup$ Are you sure that the norm is 2? Doesn't the function $f(x) = \sin(4\pi x)$ suggest that the norm is 1? $\endgroup$ Dec 14, 2015 at 18:09
  • $\begingroup$ Did you show the boundedness of the operator using sequences of functions? $\endgroup$
    – user110320
    Oct 23, 2017 at 0:22

3 Answers 3

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Consider the functions $f_n$ given by $f_n(x)=-1$ if $x\leq 1-1/n$ and $f_n(x)=2nx-2n+1$ if $1-1/n\leq x\leq 1$ (i.e. the linear interpolation between $(1-1/n,-1)$ and $(1,1)$). Then we have $$Tf_n(x)=f_n(x)-\int_0^1 f_n(s) ds=f_n(x)+1-1/n$$ and $\|Tf_n\|=2-1/n$. So indeed $\|T\|=2$.

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  • $\begingroup$ I've seen this technique applied in another problem before. As naive as this might sound, how do you know the form that this sequence of functions should take? That is, given we want to show that $\|T\|=2$, how did you know to set $(f_n)\subset X$ as you did? $\endgroup$ Dec 14, 2015 at 20:26
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    $\begingroup$ That basicly comes with a lot of practice and some intuition. In an exercise like this, where you already guessed what the norm should be, you look at the form of the operator. Here you want a function with sup 1 and integral "close to" -1... $\endgroup$
    – This Is Me
    Dec 14, 2015 at 21:27
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Let $f$ be any continuous real function such that $$ \inf_{x\in[0,1]}f(x)= -1,\;\;\;\sup_{x\in [0,1]}=1 $$ Then $\|f\|=1$. If the integral average $\int_{0}^{1}f(x)dx$ of $f$ is anything other than $0$, then $\|f\| < \|f-\int_{0}^{1}f(t)dt\|$. You can imagine that it is possible to choose $f$ so that the integral average is arbitrarily close to $1$, even thought $f$ assumes values of $\pm 1$; for such a function $Tf=f-\int_{0}^{1}f(t)dt$ will have a norm that can be made arbitrarily close to $2$.

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Consider the function $$\begin{array}{l|rcll} f_n : & [0,1] & \longrightarrow & \mathbb R \\ & x & \longmapsto & 1 - 2nx & \text{ for } 0 \le x \le \frac{1}{n}\\ & x & \longmapsto & -1 & \text{ for } \frac{1}{n} < x \le 1 \end{array}$$

You have $\Vert f_n \Vert = 1$ for all $n \in \mathbb N$ and $$\int_0^1 f_n(t) dt=-(1-\frac{1}{n})$$ Hence $$T (f_n)(x)=f_n(x)+\left(1-\frac{1}{n}\right)$$ In particular $T(f_n)(0)=2-\frac{1}{n}$ tends to $2$ as $n \to +\infty$. Which implies that $\lim\limits_{n \to +\infty} \Vert T(f_n) \Vert = 2$ and $\Vert T \Vert = 2$

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