Can anybody help me with this problem of analysis?

Question

Let $X = \{f ∈ C^2([0,3]); f(3) = 0\}$.
I have to check if $$\Vert x\Vert = \max\{\sup_{x∈[0,3]}\vert f'(x)\vert, \sup_{x∈[0,3]}\vert f''(x)\vert\}$$ is the norm in $X$ and compute the norm of the vector $f(x) = xe^{−2x}$ in $X$ and in $L^2(0,∞)$.

Answer 1

If $\|x\|_0$ then $\sup_{x\in[0,3]}|f'(x)|=0$. Together with $f(3)=0$, this implies that $f$ is identically zero. Conversely, if $f$ is the zero function, the $f'=f''=0$, so that $\|f\|=0$. This establishes positive definiteness.

For $f\in X$ and $c\in \mathbb C$ we have \begin{align} \|cf\| &= \max\left\{\sup_{x\in[0,3]}|cf'(x)|,\sup_{x\in[0,3]}|cf''(x)| \right\}\\ &= \max\left\{|c|\sup_{x\in[0,3]}|f'(x)|,|c|\sup_{x\in[0,3]}|f''(x)| \right\}\\ &= |c|\max\left\{\sup_{x\in[0,3]}|f'(x)|,\sup_{x\in[0,3]}|f''(x)| \right\}\\ &= |c|\|f\|. \end{align} This satisfies homogeneity.

For $f,g\in X$ we have \begin{align} \|f+g\| &= \max\left\{\sup_{x\in[0,3]}|(f+g)'(x)|,\sup_{x\in[0,3]}|(f+g)''(x)| \right\}\\ &\leqslant \max\left\{\sup_{x\in[0,3]}|f'(x)|+\sup_{x\in[0,3]}|g'(x)|, \sup_{x\in[0,3]}|f''(x)|+ \sup_{x\in[0,3]}|g''(x)| \right\}\\ &\leqslant \max\left\{\sup_{x\in[0,3]}|f'(x)|, \sup_{x\in[0,3]}|f''(x)| \right\} + \max\left\{\sup_{x\in[0,3]}|g'(x)|, \sup_{x\in[0,3]}|g''(x)| \right\}\\ &=\|f\| + \|g\|, \end{align}

which satisfies the triangle inequality. We conclude that $\|\cdot\|$ is a norm on $X$.

We compute the norm of $f(x) = xe^{-2x}$ in $X$ by considering its derivatives: \begin{align} f'(x) &= e^{-2x}(1-2x)\\ f''(x) &= 4e^{-2x}(x-1)\\ f'''(x) &= 4e^{-2x}(3-2x)\\ &\ \ \ \vdots\\ f^{(n)}(x) &= x^{e^{-2 x}-n} \left(-n+e^{-2 x}+1\right){}_n. \end{align} Clearly $f\in C^\infty([0,1])$, so in particular $f'$ is continuous on $[0,1]$ and differentiable on $(0,1)$. It follows that a global maximum of $|f'|$ must occur on a boundary point or a critical point. We compute \begin{align} |f'(0)| &=\left| e^{-2\cdot 0}(1-2\cdot 0) \right| = 1,\\ |f'(1)| &= \left| e^{-2\cdot 1}(1-2\cdot 1)\right| = |-e^{-2}| = e^{-2}, \end{align} and setting $f''(x)= 0$, we have $x=1$. It follows that $\sup_{x\in[0,3]}|f'(x)| = |f'(0)|=1$.

Similar analysis on $f''$ yields \begin{align} |f''(0)| &= \left|4e^{-2\cdot 0}(0-1) \right| = 4\\ |f''(1)\ &= \left|4e^{-2\cdot 1}(1-1) \right| = 0, \end{align} and setting $f'''(x) = 0$, we have $x=\frac32$. Evaluating the absolute value of $f''$ at $x=\frac32$, we have $$\left|f''\left(\frac32\right)\right| = \left| 4e^{-2\cdot\frac 32}\left(1-2\cdot\frac 32\right)\right|= 8e^{-3}.$$ It follows that $\sup_{x\in[0,3]}|f''(x)| = |f''(0)| = 4$. Therefore $$ \|xe^{-x}\| = \max\left\{\sup_{x\in[0,3]}|f'(x)|, \sup_{x\in[0,3]}|f''(x)|\right\} = 4. $$

Finally, we compute the $L^2$ norm of $f$ on $(0,\infty)$ by $$ \sqrt{\int_0^{\infty } \left|x e^{-2 x}\right|^2 \, dx} = \sqrt\frac1{32} = \frac1{4\sqrt 2}. $$