Piecewise function in $L^p$ spaces

Consider the space $$C[0,3]$$ for piecewise function such that $$f_a(x)= \begin{cases}a^3(2-a^3x),& 0\le x \le \frac2{a^3}\quad\text{and} \\[1ex] 0 , & \frac2{a^3}\le x \le 3\end{cases}$$

Set $$\;B=\{f_a(x) : a\ge 1 \}$$. Show that

1. $$B$$ is not bounded in $$(L([0,3]) ,\| \cdot\|_\infty)$$
2. $$B$$ is bounded in $$(L([0,3]) ,\| \cdot\|_1)$$
3. B is not bounded in $$(L([0,3]) ,\| \cdot\|_2)$$

I think about the question, but I didn't understand it. I get $$\sup ( {f_1, f_2, ...} )= 2$$ in the $$\|\cdot\|_\infty$$. Naturally, there is only one answer and it is not bounded this is ok. But what about the others? How bounded can we say in $$\||\cdot\|_1$$? The value of the integral proceeds independently of the value $$a$$, and the result is $$2$$. So $$B ({f_1 + f_2 ...}) = B ({2 + 2 ....}$$ ) continues. I think the result goes to infinity. I don't know how it is bounded . I checked its graph. As $$x$$ approaches $$0$$, the function goes to infinity. I'm really confused. I need your help.

$$B$$ is NOT the sum of the integrals, its just the collection of functions $$f_a$$.
Remember that a set is bounded if there is a ball of finite radius that contains the whole set, since each integral in $$B$$ integrates to $$4 - 2/a^3$$ and $$2/a^3 \leq 2$$ for $$a \geq 1$$ just pick any $$f_a$$ and the ball of radius $$8$$ at this point covers $$B$$, since for any $$f_k$$ in $$B$$
$$\lvert\lvert f_k - f_a \rvert\rvert_1 \leq \lvert\lvert f_k \rvert\rvert_1 + \lvert\lvert f_a \rvert\rvert_1 \leq (4 - 2/a^3) + (4 - 2/a^3) \leq 8$$
• @Dore $a$ is constant, the integral is $2a^3 x - \dfrac{a^3x^2}{2}$ the exponent on your $a^6$ is wrong. – Aram May 6 at 3:29