Let $C_0([0, 1])$be a subspace of $C([0, 1])$, a functional space consisting of real-value continuous functions over the interval $[0, 1]$, such that
$C_0 ([0, 1]) = \left\{ f \in C([0, 1]) \mid \int_0^1 f(t) dt = 0 \right\}$
, and define the norm as $\| f \|_\infty = \sup_{x \in [0, 1]} |f(x)|$.
Then, define linear operator $L: C_0 ([0, 1]) \to C ([0, 1])$ as
$(Lf)(x) = \int_0^x (x-t)f(t)\, dt\quad (x \in [0, 1])$
I can show that $L$ is bounded by using some inequalities, but what is the operator norm $||L||$?
So far, I hypothesize that $\|L\| = \frac{1}{4}$, by considering the definition $\|L\| = \sup\{\|Lf\|_\infty: f \in C_0 ([0, 1]) {\rm with} \|f\|_\infty = 1\}$, and then thinking of a continuous function that is very close to this one:
$f(x) = \left\{ \begin{array}{ll} 1 & (0 \leq x \leq 1/2) \\ -1 & (1/2 < x \leq 1) \end{array} \right.$
(I know this is not even continuous, but I'm thinking of an intuitive way to estimate $\|L\|$ by thinking of a function $f$ that satisfies $\|f\|_\infty = 1$, and would give the maximum of $\|Lf\|_\infty$.)
And then I get the $\frac{1}{4}$by calculating (assume $x > 1/2$)
$\int_0^x (x-t)f(t)\, dt = \int_0^{1/2} (x - t)\, dt + \int_{1/2}^x (t-x)\, dt = -\frac{1}{2}x^2 + x -\frac{1}{4}$
and finding the maximum value of the result ($\frac{1}{4}$ at $x = 1$)
Where do I go from here? How can I give a more mathematical approach to calculating $\|L\|$? Thank you in advance.
Hints: Let $f_n(x)=-n(x-\frac 1 2 +\frac 1 n)+1$ for $\frac 1 2 -\frac 1 n \leq x \leq \frac 1 2 +\frac 1 n$, $1$ for $x \leq \frac 1 2 -\frac 1 n$ and $-1$ for $x \geq \frac 1 2 +\frac 1 n$. Then $f_n$ is continuous and $\int f_n(x)dx=0$. Observe also that ($Lg$ can be defined for any integrable function $g$) and $|Lf(x)| \leq \int|f(t)|dt$ for all $x$. Show that $\int |f_n(x)-f(x)|dx \to 0$ (where $f$ is the discontinuous function you have introduced). Conclude that $Lf_n \to Lf$. Can you finish?