Difference between revisions of "Talk:Aufgaben:Problem 15"

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Let \(t>0\) fixed. Define \( f(x) := x^{t-1}e^{-x} \), non-negative function on (0,\(\infty\)).
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--[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 11:38, 11 June 2015 (CEST)
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==Part a)==
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Just a remark. Wikipedia says \(T(\phi)\) continous \( \Leftrightarrow \exists k\in \mathbb{N}_0 \;|T(\phi)| \leq C \|\phi\|_{C_b^k(K)} := C \sum_{|\alpha| \leq k}^{} \sup_{x \in K}\left| \partial^\alpha \phi(x) \right|\)
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===Problem===
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For \(k = 1,2,... \) let $$ f_k(x) = \left\{ \begin{array}{l l} x^{t-1}(1-\frac{x}{k})^{k} & \quad 0<x<k\\ 0 & \quad k\leq x<\infty\end{array} \right. $$
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That would probably relay more on part a) than your solution, but what you did, looks fine too to me.
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Show that \(f_k(x) \rightarrow f(x) \) and \(f_k(x) \leq f_{k+1}(x), \forall x>0.\)
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''Hint'': use the arithmetic-geometric mean inequality. You may ''not'' use the definition of the exponential function as a limit.
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--[[User:yzemp|yzemp]] ([[User talk:yzemp|talk]]) 14:19, 11 June 2015 (CEST)
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Yeah, we were aware of this definition, the problem is that we can't rely on wikipedia definitions when we solve these tasks. We found it better to stick to the script, this way we can be sure that we are allowed to use the definitions.
  
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===Solution===
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--[[User:Djanine|Djanine]] ([[User talk:Djanine|talk]]) 14:25, 11 June 2015 (CEST)
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Besides, we wanted to solve it as easily and short as possible.
  
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'''To show:'''
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----
  
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$$ \lim_{k \to \infty} f_{k}(x) = f(x), \forall x>0 $$
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In series 4 we had a similar exercise as in (b )there we needed to use the uniform convergence of \(\phi_j\) to show that \(\delta\) is a distribution. I remember using pointwise convergence which was marked as incorrect. Anyone know why pointwise convergence is not enough? I would think we need to do the same here. Edit: I think especially we need to consider the case where \(B\cap K = \emptyset\) separately. I don't think using only pointwise convergence is a problem. But it is convenient to use uniform convergence considering (a).
  
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'''Proof:'''
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In both parts it is probably enough to prove (2) and (4) and then set f equal to one.
  
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Take any \( x \in (0 ; \infty) \) and \( k > x \) ; consider:
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 16:31, 30 July 2015 (CEST)
  
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$$ \ln((1-\frac{x}{k})^{k}) = k\ln(1-\frac{x}{k}) = \frac{x}{x}k(\ln(1-\frac{x}{k})-\ln(1)) = -x\dfrac{(\ln(1)-\ln(1-\frac{x}{k}))}{\frac{x}{k}} $$
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----
  
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=== Alternative solution of the continuousness in part b) ===
  
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The last fraction is the difference quotient of ln evaluated in 1, thus:
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Wikipedia says: a linear functional \(T: \mathcal D (\Omega) \to \mathbb{C} \) is a distribution \( \Leftrightarrow \forall \) compact subspaces \( K \subset \Omega \exists k\in \mathbb{N}_0 \) and \( \exists C>0 \) such that \( |T(\phi)| \le C \|\phi\|_{C_b^k(K)} := C \sum\limits_{|\alpha| \leq k}^{} \sup\limits_{x \in K} |\partial^\alpha \phi(x)|\).
  
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$$ \lim_{k \to \infty} \ln((1-\frac{x}{k})^{k}) = -x\dfrac{d}{dy}\ln(y)\mid_{y=1} = -x $$
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For \( k = 0\) this becomes \( C \sup\limits_{x \in K} |\phi(x)|\).
  
  
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''An alternative version would be:'' \( \lim_{k \rightarrow \infty} \ln((1-\frac{x}{k})^{k}) = \lim_{k \rightarrow \infty}  k\ln(1-\frac{x}{k}) = \lim_{k \rightarrow \infty}  \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} \) and then use Bernoulli-L'Hopital to get: \( \lim_{k \rightarrow \infty}  \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} = \lim_{k \rightarrow \infty}  \frac{\frac{x}{k^2}}{(\frac{x}{k} - 1) \frac{1}{k^2}} = -x \)
 
  
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In part a) we have shown that \( |1_B(\phi)| = |\langle 1_B,\phi \rangle| \le |B| \sup\limits_{x \in B}|\phi (x)| \enspace(\le |B| \sup\limits_{x \in \mathbb{R}^3}|\phi (x)| )\).
  
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Since exp is continuous, we can "bring the limes inside":
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If we can show that \(|B| \sup\limits_{x \in B}|\phi (x)| \le C \sup\limits_{x \in K} |\phi(x)| \enspace \forall \) compact subspaces \( K \subset \mathbb{R}^3 \), the exercise will be done.
  
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$$ \Rightarrow \lim_{k \to \infty} f_{k}(x) =  x^{t-1} \lim_{k \to \infty} \exp ( \ln((1-\frac{x}{k})^{k}) = x^{t-1} \exp(\lim_{k \to \infty} ln((1-\frac{x}{k})^{k}) = x^{t-1} e^{-x} = f(x) $$
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$$ \Rightarrow C \ge^! \frac{|B| \sup\limits_{x \in B}|\phi (x)|}{\sup\limits_{x \in K}|\phi (x)|} $$
  
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\( \blacksquare \)
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As \(\sup\limits_{x \in K}|\phi (x)| \neq 0\) and \(|B| \sup\limits_{x \in B}|\phi (x)| < \infty \), we can find a finite \(C\) to fullfil this equation.
  
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Thus it follows that \( 1_B\) is a distribution.
  
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'''To show:'''
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\(\sup\limits_{x \in K}|\phi (x)| \neq 0\) not sure about that.
  
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$$ f_{k}(x) \leq f_{k+1}(x), \forall x > 0 $$
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---------------------------------------------------------------------------------------
  
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'''Proof:'''
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Correct me, but I think this step is wrong. The absolute value should be taken inside the supremum.
  
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''Case 1:'' \( x \in \: ]0;k[ \)
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$$ \left| \int_{B}\phi(x)dx \right| \le \left|  |B| \sup_{x \in B}(\phi (x)) \right| $$
  
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The arithmetic-geometric mean equality says that \( \forall x_{1},...,x_{n} \geq 0 \) it holds: $$ (x_{1}...x_{n})^{\frac{1}{n}} \leq \dfrac{x_{1}+...+x_{n}}{n} \tag{$\dagger$}$$
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[[User:Beni|Beni]] ([[User talk:Beni|talk]]) 15:38, 3 August 2015 (CEST)
  
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Consider:
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I agree, shouldn't it be:
  
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$$ ((1-\dfrac{x}{k})^{k}\cdot1)^{\frac{1}{k+1}} \overset{(\dagger)}{\leq} \dfrac{1}{k+1}(k(1-\dfrac{x}{k})+1) = \dfrac{1}{k+1}(k + 1 - x) = (1 - \dfrac{x}{k+1}) $$
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$$ \left| \int_{B}\phi(x)dx \right| \leq  \int_{B} |\phi(x)|dx \leq   |B| \sup_{x \in B} |\phi(x)| \leq \dots$$
  
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$$\Rightarrow (1-\dfrac{x}{k})^{k} \leq (1 - \dfrac{x}{k+1})^{k+1}  $$
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Also is nobody with me on (b)? (see above)
  
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Where we don't have any problem with the k+1-root because \( (1-\dfrac{x}{k})^{k} \) only takes real positive values, since \( 0 < x < k. \)
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 16:38, 3 August 2015 (CEST)
  
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''Case 2:'' \( x \in [k;k+1[ \)
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Jep, that seems correct. I think we should use the definition as it is in the lecture notes. Therefore argue with \( \phi_j(x) \) uniformally convergent and since according to a) the supremum of this sequence exits which gives us the upper bound to apply lebesgue dominated convergence.
  
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The inequality holds since \( f_{k}(x) = 0 \) and \( f_{k+1}(x) = x^{t-1}(1 - \dfrac{x}{k+1})^{k+1} \) is a positve function \( \forall  x \in [k;k+1[. \)
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[[User:Beni|Beni]] ([[User talk:Beni|talk]]) 16:49, 3 August 2015 (CEST)
  
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''Case 3:'' \( x \in [k+1;\infty[ \)
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Yes I would agree if \(\{\phi_j\}\) would uniformly converge on \(\mathbb{R}^3\). But have you seen the definition of uniform convergence in \(\mathcal{D}(\Omega)\) given in the script? The sequence only converges on \(K\). Also take a look at the ML of series 4 ex.2(a).
  
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Both functions are equal zero so the inequality still holds.
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 17:05, 3 August 2015 (CEST)
  
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\( \blacksquare \)
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Just as given on page 8 on Newtonian Potential for any compact subset that contains \(sup(\phi_j) \)
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I think the definition is as usual that in the limes the difference of phi_j and phi goes to zero.
  
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==Part b)==
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[[User:Beni|Beni]] ([[User talk:Beni|talk]]) 17:31, 3 August 2015 (CEST)
  
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===Problem===
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Not sure what you mean. The way I understand this is that \(K\) is fixed for a given converging series \(\{\phi_j\}\) and therefore we can't just assume \( B \subset K\). Then \(\lim\limits_{j\rightarrow \infty} \phi_j(x) = \phi(x)\) as used in the current solution is strictly speaking not correct.
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Use the monotone convergence theorem to show that $$ \Gamma(t) = \lim_{k \to \infty} \frac{k!k^{t}}{t(t+1)...(t+k)}, t>0 $$
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where \( \Gamma(t) \) is the Euler Gamma function.
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''Hint'': consider \( I_{k} = k^{t} \int_0^1 \! u^{t-1}(1-u)^{k}du \).
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 18:36, 3 August 2015 (CEST)
  
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===Solution===
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Ok, I see my mistake(s) now.
  
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'''Idea:'''
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 08:13, 4 August 2015 (CEST)
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In a first step we show that \( I_{k} \) is equal to the fraction in the right hand side of the equality. Then we rewrite \( I_{k} \) in a more convenient way, looking for something containing the functions given in part a). Eventually we let \( k\rightarrow\infty \) and apply Lebesgue's monoton convergence theorem to conclude.
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$$ I_{k} = k^{t} \int_0^1 \! u^{t-1}(1-u)^{k} \, \mathrm{d}u \overset{P.I.}{=} \\
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Though I still think it is saver to do this:
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=k^{t}\lbrace\dfrac{u^{t}}{t}(1-u)^{k}\vert_{0}^{1} - \int_0^1 \! \dfrac{u^{t}}{t}(-1)k(1-u)^{k-1} \, \mathrm{d}u \rbrace = \\
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= k^{t}\lbrace 0 + \dfrac{k}{t}\int_0^1 \! u^{t}(1-u)^{k-1} \, \mathrm{d}u \rbrace =\\
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= \dfrac{k^{t+1}}{t}\int_0^1 \! u^{t}(1-u)^{k-1} \, \mathrm{d}u \overset{P.I.}{=}\\
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=\dfrac{k^{t+1}}{t}\lbrace\dfrac{u^{t+1}}{t+1}(1-u)^{k-1}\vert_{0}^{1} - \int_0^1 \! \dfrac{u^{t+1}}{t+1}(-1)(k-1)(1-u)^{k-2} \, \mathrm{d}u \rbrace = \\
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= \dfrac{k^{t+1}}{t}\lbrace 0 + \dfrac{k-1}{t+1}\int_0^1 \! u^{t+1}(1-u)^{k-2} \, \mathrm{d}u \rbrace =\dfrac{k^{t+1}(k-1)}{t(t+1)} \int_0^1 \! u^{t+1}(1-u)^{k-2} \, \mathrm{d}u \overset{P.I.}{=}\\
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= ... = \frac{(k-1)!k^{t+1}}{t(t+1)...(t+k-1)} \int_0^1 \! u^{t+k-1} \, \mathrm{d}u =  \frac{(k-1)!k^{t+1}}{t(t+1)...(t+k-1)}\dfrac{u^{t+k}}{t+k}\vert^{1}_{0} = \frac{k!k^{t}}{t(t+1)...(t+k)} $$
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Where we have used k-1 times partial integration to reach the last but one step.
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\(f\cdot 1_B(\phi) := \langle f\cdot 1_B, \phi \rangle = \int_{\mathbb{R}^3} f\cdot 1_B(x)\phi (x) dx  \)
  
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$$  I_{k} = k^{t} \int_0^1 \! u^{t-1}(1-u)^{k} \, \mathrm{d}u =  k^{t-1} \int_0^k \! (\dfrac{x}{k})^{t-1}(1-\dfrac{x}{k})^{k} \, \mathrm{d}x = \\
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We show that \(f\cdot 1_B\) is a distribution on \(\mathcal{D}(\mathbb{R}^3)\), that is a continues, linear functional. Linearity is clear from the linearity of the integral.
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=  \int_0^k \! x^{t-1}(1-\dfrac{x}{k})^{k} \, \mathrm{d}x = \int_0^k \! x^{t-1}(1-\dfrac{x}{k})^{k} \, \mathrm{d}x + \int_k^\infty \! 0 \, \mathrm{d}x = \int_0^k \! f_{k}(x) \, \mathrm{d}x + \int_k^\infty \! f_{k}(x) \, \mathrm{d}x =\\
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= \int_0^\infty \! f_{k}(x) \, \mathrm{d}x $$
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Where the first step is a change of variable \( u = \dfrac{x}{k} \).
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For any converging sequence \(\{\phi_j\}\) in \(\mathcal{D}(\mathbb{R}^3)\) there is a compact \(K\subset \mathbb{R}^3\) such that supp\((\phi_j) \subset K\) and \(\{\phi_j\}\) converges uniformly on \(K\), or \(\lim\limits_{j\rightarrow \infty}  \sup\limits_{x\in K}|\phi_j(x)- \phi(x)| = 0\).
  
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Letting \( k \rightarrow\infty \), using Lebesgue's monoton convergence theorem on \( f_{k}(x) \), we obtain:
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$$\lim\limits_{j\rightarrow \infty}  |f\cdot 1_B(\phi_j) - f\cdot 1_B(\phi)| = \lim\limits_{j\rightarrow \infty} | \langle f\cdot 1_B, \phi_j-\phi \rangle |$$
  
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$$ I_{\infty} =  \int_0^\infty \! x^{t-1}e^{-x}\, \mathrm{d}x = \Gamma(t) $$
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$$ \overset{(a)}{\leq} \lim\limits_{j\rightarrow \infty} |B|\ {||f||}_{L^\infty(B)}  \sup\limits_{x\in \mathbb{R}^3}|\phi_j(x) -\phi(x)| =  |B|\ {||f||}_{L^\infty(B)}  \lim\limits_{j\rightarrow \infty} \sup\limits_{x\in K}|\phi_j(x) -\phi(x)| = 0$$
  
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Where the conditions to apply Lebesgue's monoton convergence theorem are satisfied (see part a), because \( f_{k}(x) \) are non-negative functions and are continuous (and therefore integrable) on \( [0;\infty[ \).
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The last step follows form the fact that for \(x\in \mathbb{R}^3\setminus K\), \(\phi_j(x) = \phi(x) = 0\). It follows that \(\lim\limits_{j\rightarrow \infty}  f\cdot 1_B(\phi_j) =  f\cdot 1_B(\phi)\).
  
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'''Lebesgue monotone convergence theorem (from exercise class, hope it's correct)'''
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This way we also directly use (a), but I don't think the other way is wrong, so it doesn't really matter.
  
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Let \( f_n \) be a sequence of integrable functions so that \( 0 \leq f_n \leq f_{n+1} \forall n \in \mathbb{N}, \forall x \in \Omega \subseteq \mathbb{R}^m \) and \( \lim_{n \to \infty} f_n(x) =: f(x) \in \mathbb{R} \: \forall x \in \Omega; \)
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 15:22, 4 August 2015 (CEST)
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then \( f \) is integrable and \( \lim_{n \to \infty} \int_\Omega f_n(x)dx = \int_\Omega f(x)dx. \)
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\( \blacksquare \)
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Thanks Carl. That looks very nice
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[[User:Beni|Beni]] ([[User talk:Beni|talk]]) 20:08, 4 August 2015 (CEST)

Latest revision as of 18:08, 4 August 2015

--Brynerm (talk) 11:38, 11 June 2015 (CEST) Just a remark. Wikipedia says \(T(\phi)\) continous \( \Leftrightarrow \exists k\in \mathbb{N}_0 \;|T(\phi)| \leq C \|\phi\|_{C_b^k(K)} := C \sum_{|\alpha| \leq k}^{} \sup_{x \in K}\left| \partial^\alpha \phi(x) \right|\)

That would probably relay more on part a) than your solution, but what you did, looks fine too to me.

--yzemp (talk) 14:19, 11 June 2015 (CEST) Yeah, we were aware of this definition, the problem is that we can't rely on wikipedia definitions when we solve these tasks. We found it better to stick to the script, this way we can be sure that we are allowed to use the definitions.

--Djanine (talk) 14:25, 11 June 2015 (CEST) Besides, we wanted to solve it as easily and short as possible.

In series 4 we had a similar exercise as in (b )there we needed to use the uniform convergence of \(\phi_j\) to show that \(\delta\) is a distribution. I remember using pointwise convergence which was marked as incorrect. Anyone know why pointwise convergence is not enough? I would think we need to do the same here. Edit: I think especially we need to consider the case where \(B\cap K = \emptyset\) separately. I don't think using only pointwise convergence is a problem. But it is convenient to use uniform convergence considering (a).

In both parts it is probably enough to prove (2) and (4) and then set f equal to one.

Carl (talk) 16:31, 30 July 2015 (CEST)

Alternative solution of the continuousness in part b)

Wikipedia says: a linear functional \(T: \mathcal D (\Omega) \to \mathbb{C} \) is a distribution \( \Leftrightarrow \forall \) compact subspaces \( K \subset \Omega \exists k\in \mathbb{N}_0 \) and \( \exists C>0 \) such that \( |T(\phi)| \le C \|\phi\|_{C_b^k(K)} := C \sum\limits_{|\alpha| \leq k}^{} \sup\limits_{x \in K} |\partial^\alpha \phi(x)|\).

For \( k = 0\) this becomes \( C \sup\limits_{x \in K} |\phi(x)|\).


In part a) we have shown that \( |1_B(\phi)| = |\langle 1_B,\phi \rangle| \le |B| \sup\limits_{x \in B}|\phi (x)| \enspace(\le |B| \sup\limits_{x \in \mathbb{R}^3}|\phi (x)| )\).

If we can show that \(|B| \sup\limits_{x \in B}|\phi (x)| \le C \sup\limits_{x \in K} |\phi(x)| \enspace \forall \) compact subspaces \( K \subset \mathbb{R}^3 \), the exercise will be done.

$$ \Rightarrow C \ge^! \frac{|B| \sup\limits_{x \in B}|\phi (x)|}{\sup\limits_{x \in K}|\phi (x)|} $$

As \(\sup\limits_{x \in K}|\phi (x)| \neq 0\) and \(|B| \sup\limits_{x \in B}|\phi (x)| < \infty \), we can find a finite \(C\) to fullfil this equation.

Thus it follows that \( 1_B\) is a distribution.

\(\sup\limits_{x \in K}|\phi (x)| \neq 0\) not sure about that.

Correct me, but I think this step is wrong. The absolute value should be taken inside the supremum.

$$ \left| \int_{B}\phi(x)dx \right| \le \left| |B| \sup_{x \in B}(\phi (x)) \right| $$

Beni (talk) 15:38, 3 August 2015 (CEST)

I agree, shouldn't it be:

$$ \left| \int_{B}\phi(x)dx \right| \leq \int_{B} |\phi(x)|dx \leq |B| \sup_{x \in B} |\phi(x)| \leq \dots$$

Also is nobody with me on (b)? (see above)

Carl (talk) 16:38, 3 August 2015 (CEST)

Jep, that seems correct. I think we should use the definition as it is in the lecture notes. Therefore argue with \( \phi_j(x) \) uniformally convergent and since according to a) the supremum of this sequence exits which gives us the upper bound to apply lebesgue dominated convergence.

Beni (talk) 16:49, 3 August 2015 (CEST)

Yes I would agree if \(\{\phi_j\}\) would uniformly converge on \(\mathbb{R}^3\). But have you seen the definition of uniform convergence in \(\mathcal{D}(\Omega)\) given in the script? The sequence only converges on \(K\). Also take a look at the ML of series 4 ex.2(a).

Carl (talk) 17:05, 3 August 2015 (CEST)

Just as given on page 8 on Newtonian Potential for any compact subset that contains \(sup(\phi_j) \) I think the definition is as usual that in the limes the difference of phi_j and phi goes to zero.

Beni (talk) 17:31, 3 August 2015 (CEST)

Not sure what you mean. The way I understand this is that \(K\) is fixed for a given converging series \(\{\phi_j\}\) and therefore we can't just assume \( B \subset K\). Then \(\lim\limits_{j\rightarrow \infty} \phi_j(x) = \phi(x)\) as used in the current solution is strictly speaking not correct.

Carl (talk) 18:36, 3 August 2015 (CEST)

Ok, I see my mistake(s) now.

Carl (talk) 08:13, 4 August 2015 (CEST)

Though I still think it is saver to do this:

\(f\cdot 1_B(\phi) := \langle f\cdot 1_B, \phi \rangle = \int_{\mathbb{R}^3} f\cdot 1_B(x)\phi (x) dx \)

We show that \(f\cdot 1_B\) is a distribution on \(\mathcal{D}(\mathbb{R}^3)\), that is a continues, linear functional. Linearity is clear from the linearity of the integral.

For any converging sequence \(\{\phi_j\}\) in \(\mathcal{D}(\mathbb{R}^3)\) there is a compact \(K\subset \mathbb{R}^3\) such that supp\((\phi_j) \subset K\) and \(\{\phi_j\}\) converges uniformly on \(K\), or \(\lim\limits_{j\rightarrow \infty} \sup\limits_{x\in K}|\phi_j(x)- \phi(x)| = 0\).

$$\lim\limits_{j\rightarrow \infty} |f\cdot 1_B(\phi_j) - f\cdot 1_B(\phi)| = \lim\limits_{j\rightarrow \infty} | \langle f\cdot 1_B, \phi_j-\phi \rangle |$$

$$ \overset{(a)}{\leq} \lim\limits_{j\rightarrow \infty} |B|\ {||f||}_{L^\infty(B)} \sup\limits_{x\in \mathbb{R}^3}|\phi_j(x) -\phi(x)| = |B|\ {||f||}_{L^\infty(B)} \lim\limits_{j\rightarrow \infty} \sup\limits_{x\in K}|\phi_j(x) -\phi(x)| = 0$$

The last step follows form the fact that for \(x\in \mathbb{R}^3\setminus K\), \(\phi_j(x) = \phi(x) = 0\). It follows that \(\lim\limits_{j\rightarrow \infty} f\cdot 1_B(\phi_j) = f\cdot 1_B(\phi)\).

This way we also directly use (a), but I don't think the other way is wrong, so it doesn't really matter.

Carl (talk) 15:22, 4 August 2015 (CEST)

Thanks Carl. That looks very nice

Beni (talk) 20:08, 4 August 2015 (CEST)

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