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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.

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$B$ 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 $B$

$$ \lvert\lvert f_k - f_a \rvert\rvert_1 \leq \lvert\lvert f_k \rvert\rvert_1 + \lvert\lvert f_a \rvert\rvert_1 \leq (4 - 2/a^3) + (4 - 2/a^3) \leq 8$$

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  • $\begingroup$ How did you get 4- (2/a^3) ? I obtained result just 2. 2xa^3-(a^6x^2)/2 if we write x=2/a^3 then result is 2. Where is my mistake? $\endgroup$ – Dore May 5 at 21:39
  • $\begingroup$ @Dore $a$ is constant, the integral is $2a^3 x - \dfrac{a^3x^2}{2}$ the exponent on your $a^6$ is wrong. $\endgroup$ – Aram May 6 at 3:29

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