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(a) $A= \{f \in C^2[0,1]:f(x)>0,\parallel f'\parallel_{\infty}<1, |f''(0)|>2\}$

(b) $B= \{f\in C^2[0,1]:f(1)<0,f'(1)=0,f''(1)>0 \}$

(c) $C= \{f\in C^2[0,1]:f(x)f'(x)>0$ for $0\le x \le 1\}$

(d) $D= \{f\in C^2[0,1]: f(x)f'(x)>0$ for $0\lt x \lt 1\}$

I have the intuition that $A$ is an open set since it includes all $f(x)>0$, $B$ is a neither a closed nor open set since it only holds true for one element in the interval, $C$ is a closed set since it includes an open set and has a boundary and $D$ is an open set since it doesn't include the boundary.

Any feed back would be appreciated,


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Which norm are you using on $C^2[0,1]$? – MichaelNgelo Feb 1 '13 at 5:25
Indeed, what topology do you use? – HSN Feb 1 '13 at 12:35
@HSN topology of the normed spaces – user5208 Feb 1 '13 at 14:45
You didn't answer the question, there are several possible norms on $C^2[0,1]$. – Seirios Feb 21 '13 at 16:50
@JonasMeyer real analysis and applications by davidson and donsig page 119 exercise E – user5208 Feb 21 '13 at 18:47
up vote 2 down vote accepted

I use the norm $||f||_{C^2}= \max\limits_{0 \leq i \leq 2} ||f^{(i)}||_{\infty}$ (introduced in example 7.1.4 of the book).

(a) $A$ is open:

Let $f \in A$, $m= \min\limits_{x \in [0,1]} f(x) >0$ and $\epsilon>0 $ such that $\epsilon<\min(m,1-||f'||_{\infty},|f''(0)|-2)$.

For $g \in B(f,\epsilon)$, $|g(x)-f(x)| \leq ||f-g||_{\infty} < \epsilon$ so $g(x)>f(x)-\epsilon \geq m-\epsilon>0$; $||f'-g'||_{\infty}< \epsilon$ so $||g'||_{\infty}<||f'||_{\infty}+ \epsilon <1$; $|f''(0)|-|g''(0)| \leq |f''(0)-g''(0)| \leq ||f''-g''||_{\infty}<\epsilon$ so $|g''(0)| >|f''(0)|-\epsilon >2$.

Thus, $g \in A$ and $B(f,\epsilon) \subset A$.

(b) $B$ is not open:

Take $f_n(x)=-1+(x-1)(x^3- \frac{3}{2} x^2+ \frac{1}{n}+ \frac{1}{2})$ and $f(x)=-1+(x-1)(x^3- \frac{3}{2} x^2+ \frac{1}{2})$. Then $f \in B$, $f_n \notin B$ and $||f-f_n||_{C^2} \leq \frac{1}{n}$.

(c) $C$ is open:

Notice that $C=C_1 \cup C_2$ with $C_1= \{ f \in C^2[0,1] \mid f>0, f'>0 \}$ and $C_2= \{ f \in C^2[0,1] \mid f<0, f'<0 \}$. So it is sufficient to show that $C_1$ and $C_2$ are open.

Let $f \in C_1$, $m_1= \min\limits_{x \in [0,1]} f(x)>0$, $m_2= \min\limits_{x \in [0,1]} f'(x)>0$ and $\epsilon>0$ such that $\epsilon< \min(m_1,m_2)$.

For $g \in B(f,\epsilon)$, $|g(x)-f(x)|< ||g-f||_{\infty}<\epsilon$ so $g(x)>f(x)-\epsilon>m_1-\epsilon>0$; $|g'(x)-f'(x)|< ||g'-f'||_{\infty}<\epsilon$ so $g'(x)>f'(x)-\epsilon>m_2-\epsilon>0$.

So $g \in C_1$ and $B(f,\epsilon) \subset C_1$. The same argument works for $C_2$.

(d) $D$ is not open.

As $D=D_1 \cup D_2$ with $D_1= \{ f \in C^2[0,1] \mid f>0, f'>0 \ \text{on} \ (0,1) \}$, $D_2= \{ f \in C^2[0,1] \mid f<0, f'<0 \ \text{on} \ (0,1) \}$ and $D_1 \cap D_1= \emptyset$, it is sufficient to show that $D_1$ is not open.

Take $f(x)=x^3$ and $f_n(x)= \left\{ \begin{matrix} 0 & \text{if} \ 0 \leq x \leq 1/n \\ (x-1/n)^3 & \text{otherwise} \end{matrix} \right.$. Then $f \in D$, $f_n \in C^2[0,1]$ and $f_n \notin D$, but $||f_n-f||_{C^2} \leq \frac{6}{n}$.

Because a normed space is always connected, neither $A$ nor $C$ are closed. Moreover, neither $B$ nor $D$ are closed. Indeed, $f \equiv 0$ is in the closure of $D$ and of $B$ but neither in $D$ nor in $B$.

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