# Bounded Linear Operator from $C_0([0,1])$ to $C([0,1])$

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, $$\|Tf\|\leq \|f\|\,\int_0^1\,(1-t)\,\text{d}t=\frac{1}{2}\,\|f\|\,.$$

• There is no equation, it is a definition for $T$, in this case $f$ is a variable.
– zwim
Jul 9 '20 at 19:11
• 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 $\|Tf\|=\frac{1}{2}$. Jul 10 '20 at 0:33
• I understand, but I want to get T’s norm. How can I do it? Jul 10 '20 at 3:20
• @eosa Show how you got the $\leq \frac{1}{2}$ bound, since it may have a clue on how to construct such an $f$. Jul 10 '20 at 4:10

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 $$f_n(x)=\left\{ \begin{array}{ll} -1&\text{if }0\leq x\leq \dfrac{1}{2}-\dfrac1{2n}\,,\\ 2n\left(x-\dfrac12\right)&\text{if }\dfrac{1}{2}-\dfrac{1}{2n} Observe that $$\|Tf_n\|_\infty=Tf_n(1)=\dfrac14-\dfrac1{12n^2}$$ for every positive integer $$n$$.