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Define $$ C_0([0,1]) := \left\{f\in C([0,1]) : \int_0^1f(t)\, \mathrm dt=0\right\}. $$ Show that $T : C_0([0,1]) \to C([0,1])$, given by $$ (Tf)(x) := \int_0^x(t-x) f(t) \, \mathrm dt, $$ defines a bounded linear operator.

I proved $\Vert T \Vert \leq \frac 1 2$, but I could not find some $f \in C_0([0,1])$ such that $\Vert f \Vert = 1$ and $\Vert T f \Vert = \frac 1 2$.

Here is how to prove $\|T\|\leq \frac12$.

Note that $$\|Tf\|=\sup_{x\in[0,1]}\,\left|\int_0^x\,(t-x)\,f(t)\,\text{d}t\right|\leq \sup_{x\in [0,1]}\,\int_0^x\,|t-x|\,\big|f(t)\big|\,\text{d}t\,.$$ Since $\big|f(t)\big|\leq \|f\|$ for all $t\in[0,1]$, we get $$\|Tf\|\leq \sup_{x\in[0,1]}\,\int_0^x\,|t-x|\,\|f\|\,\text{d}t\leq \sup_{x\in[0,1]}\,\|f\|\,\int_0^x\,(x-t)\,\text{d}t\,.$$ Therefore, $$\|Tf\|\leq \|f\|\,\sup_{x\in[0,1]}\,\int_0^x\,(x-t)\,\text{d}t\leq \|f\|\,\sup_{x\in [0,1]}\,\int_0^x\,(1-t)\,\text{d}t\,.$$ So, $$\|Tf\|\leq \|f\|\,\int_0^1\,(1-t)\,\text{d}t=\frac{1}{2}\,\|f\|\,.$$

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  • $\begingroup$ There is no equation, it is a definition for $T$, in this case $f$ is a variable. $\endgroup$
    – zwim
    Jul 9 '20 at 19:11
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    $\begingroup$ Well if you want to show that $T$ is bounded, it's enough to check that $\|T\|\leq \frac{1}{2}$. Also, it might be that the supremum is never attained. That is, even if $\|T\|=\frac{1}{2}$, it might be that there is no $f$ with $\|Tf\|=\frac{1}{2}$. $\endgroup$ Jul 10 '20 at 0:33
  • $\begingroup$ I understand, but I want to get T’s norm. How can I do it? $\endgroup$ Jul 10 '20 at 3:20
  • $\begingroup$ @eosa Show how you got the $\leq \frac{1}{2}$ bound, since it may have a clue on how to construct such an $f$. $\endgroup$
    – lisyarus
    Jul 10 '20 at 4:10
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Let $f\in \mathcal{C}_0\big([0,1]\big)$. Fix $x\in[0,1]$. Note that $$-Tf(x)=\int_0^x\,(x-t)\,f(t)\,\text{d}t=\int_0^x\,\int_t^x\,f(t)\,\text{d}s\,\text{d}t=\int_0^x\,\int_0^s\,f(t)\,\text{d}t\,\text{d}s\,.$$ Write $$If(x):=\int_0^x\,f(t)\,\text{d}t\,.$$ Then, $$-Tf(x)=\int_0^x\,If(s)\,\text{d}s\,.$$ Observe that $$\big|If(x)\big|\leq \int_0^x\,\big|f(t)|\,\text{d}t\leq \|f\|_\infty\,x$$ and $$\big|If(x)\big|=\big|If(1)-If(x)\big|=\left|\int_x^1\,f(t)\,\text{d}t\right|\leq \|f\|_\infty\,(1-x)\,.$$ Thus, $$\big|If(x)\big|\leq \|f\|_\infty\,\min\{x,1-x\}\,.$$ Ergo, $$\big|Tf(x)\big|\leq \int_0^x\,\big|If(s)\big|\,\text{d}s\leq \|f\|_\infty\,\int_0^x\,\min\{s,1-s\}\,\text{d}s\,.$$ This implies $$\big\|Tf\big\|_\infty\leq \|f\|_\infty\,\int_0^1\,\min\{s,1-s\}\,\text{d}s=\frac{1}{4}\,\|f\|_\infty\,.\tag{*}$$ Therefore, $\|T\|_\text{op}\leq \dfrac14$.

Note that (*) is an equality if and only if $f\equiv 0$, since the equality case occurs only when there exists $\epsilon$ such that $|\epsilon|=1$ and $If(x)=\epsilon\,\|f\|_\infty\,\max\{x,1-x\}$ for all $x\in[0,1]$. This condition, along with the fact that $f$ is continuous, implies that $f\equiv 0$ is the only possible choice. However, we can find a sequence $\left(f_n\right)_{n=1}^\infty$ of functions $f_n\in\mathcal{C}_0\big([0,1]\big)$ satisfying $\|f_n\|_\infty=1$ and $\lim\limits_{n\to\infty}\,\|Tf_n\|_\infty=\dfrac14$. For each $n=1,2,3,\ldots$, take $$f_n(x)=\left\{ \begin{array}{ll} -1&\text{if }0\leq x\leq \dfrac{1}{2}-\dfrac1{2n}\,,\\ 2n\left(x-\dfrac12\right)&\text{if }\dfrac{1}{2}-\dfrac{1}{2n}<x<\dfrac12+\dfrac1{2n}\,,\\ +1&\text{if }\dfrac12+\dfrac1{2n}\leq x\leq 1\,. \end{array} \right.$$ Observe that $$\|Tf_n\|_\infty=Tf_n(1)=\dfrac14-\dfrac1{12n^2}$$ for every positive integer $n$.

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