Finding the norm of the operator $A: L^1(0,3) \rightarrow L^1(0,3)$ (Af)(x) = (x+5)f(x)

Question

Let the operator $A: L^1(0,3) \rightarrow L^1(0,3)$ be given by the formula (Af)(x) = (x+5)f(x)

Find the norm of A.

Some thoughts:

The norm of A is $inf{M>0: ||A(f)|| \leq M||f|| \forall f\in L^1(0,3)} = \underset{||f||=1}{\mathrm{sup}} \frac{||A(f)||}{||f||}$ I have seen that $||A|| \leq$ 8 as $||(x+5)f(x)|| = \int_0^3 (x+5)|f(x)| \leq 8 \int_0^3|f(x)| = 8||f||$

So if i found an ||f||=1 such that ||(x+5)f(x)||=8 I could conclude that ||A|| is 8, but I cannot find it

(maybe ||A|| is less than 8 but it must be greater than at least 7.5 because if we take ||g|| such as [g=1 if x\in[2,3], 0 otherwise] then ||A(g)||=7.5)

Answer 1

Let $f_n=n(3-1/n, 3)$. Then we have that $$\lVert f_n\rVert=\int_{3-1/n}^{3}n=1$$ And $$\lVert Af_n \rVert=\int_{3-1/n}^{3}n(x+5)\mathrm{d}x=8-\frac{1}{2n} \to 8$$ But this means that $\forall \varepsilon > 0$ we have that $\lVert A\rVert \geqslant 8 - \varepsilon$.