The operator norm ∥L∥$\Vert L\Vert$

Let C0([0,1])$C_0([0,1])$be a subspace of C([0,1])$C([0,1])$, a functional space consisting of real-value continuous functions over the interval [0,1]$[0,1]$, such that

C0([0,1])={f∈C([0,1])∣∫10f(t)dt=0}$C_0([0,1])=\{f\in C([0,1])\mid {\int }_0^{1}f(t)dt=0\}$

, and define the norm as ∥f∥∞=supx∈[0,1]|f(x)|$\Vert f{\Vert }_{\mathrm{\infty }}=\underset{x\in [0,1]}{sup}|f(x)|$.

Then, define linear operator L:C0([0,1])→C([0,1])$L:C_0([0,1])\to C([0,1])$ as

(Lf)(x)=∫x0(x−t)f(t)dt(x∈[0,1])$(Lf)(x)={\int }_0^x(x-t)f(t)dt(x\in [0,1])$

I can show that L$L$ is bounded by using some inequalities, but what is the operator norm ||L||$||L||$?

So far, I hypothesize that ∥L∥=14$\Vert L\Vert =\frac{1}{4}$, by considering the definition ∥L∥=sup{∥Lf∥∞:f∈C0([0,1])with∥f∥∞=1}$\Vert L\Vert =sup\{\Vert Lf{\Vert }_{\mathrm{\infty }}:f\in C_0([0,1])\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\Vert f{\Vert }_{\mathrm{\infty }}=1\}$, and then thinking of a continuous function that is very close to this one:

f(x)={1−1(0≤x≤1/2)(1/2<x≤1)$f(x)=\{\begin{array}{ll}1& (0\le x\le 1/2)\\ -1& (1/2<x\le 1)\end{array}$

(I know this is not even continuous, but I'm thinking of an intuitive way to estimate ∥L∥$\Vert L\Vert$ by thinking of a function f$f$ that satisfies ∥f∥∞=1$\Vert f{\Vert }_{\mathrm{\infty }}=1$, and would give the maximum of ∥Lf∥∞$\Vert Lf{\Vert }_{\mathrm{\infty }}$.)

And then I get the 14$\frac{1}{4}$by calculating (assume x>1/2$x>1/2$)

∫x0(x−t)f(t)dt=∫1/20(x−t)dt+∫x1/2(t−x)dt=−12x2+x−14${\int }_0^x(x-t)f(t)dt={\int }_0^{1/2}(x-t)dt+{\int }_{1/2}^x(t-x)dt=-\frac{1}{2}{x}^2+x-\frac{1}{4}$

and finding the maximum value of the result (14$\frac{1}{4}$ at x=1$x=1$)

Where do I go from here? How can I give a more mathematical approach to calculating ∥L∥$\Vert L\Vert$? Thank you in advance.