Math Challenge - February 2020

  • Challenge
  • Thread starter fresh_42
  • Start date
  • Featured
  • #101
152
71
@fresh_42 I think I have the solution using the hint. That hint is quite a good idea, it's probably useful to remember in general.
First, we use Riesz representation theorem to define a mapping ##T: \mathcal{H}^* \rightarrow \mathcal{H}##, which defines the mapping from the dual space of linear functionals to Hilbert space:
$$F(x) = (T(F),x)$$
for every ##x \in \mathcal{H}##, where ##(\cdot,\cdot)## is the inner product. The theorem itself states that the vector ##T(F)## exists and is unique, thus defining the mapping above. Taking into account that inner product is linear(in a real Hilbert space), we find that ##T## is also linear, namely:
$$(T(F+G),x) = (F+G)(x) = F(x)+G(x) = (T(F)+T(G),x)$$
Also, we define the mapping ##B##:
$$B(f)(h) = \beta(f,h)$$
which is possible for continuous bilinear forms. ##B## is linear by definition.
Now we induce the norm on our Hilbert space from the inner product, and metric from the norm naturally:
$$\lVert x \rVert_\mathcal{H} \equiv \sqrt{(x,x)} \qquad d_\mathcal{H}(x,y) \equiv \lVert x-y\rVert_\mathcal{H}$$
We define the mapping which was hinted(there's a typo in the hint though, I assume):
$$Q(f) \equiv f -\lambda(T(B(f)) - T(F))$$
where ##\lambda## is a free parameter.
If it is possible to find ##\lambda## such that this function has a fixed point ##x##, then we have for that point:
$$Q(x) = x \Leftrightarrow T(B(x)) = T(F)$$
and this would be equivalent to the solution of our exercise, since by Riesz we certainly have unique vectors ##T(B(f))## and ##T(F)## such that, for some ##f \in \mathcal{H}##:
$$(\forall h \in \mathcal{H})( F(h) = (T(F),h) \wedge \beta(f,h) = (T(B(f)),h))$$
but the above then proves that we have a unique vector ##f## such that ##\beta(f,h) = F(h)## for all ##h \in \mathcal{H}##. hence it proves the theorem.

So, we aim to prove that ##Q(f)## has a unique fixed point, which we will prove using Banach fixed point theorem. We first want ##Q(f)## to be a contraction.
A contraction is a function ##f## on a metric space, such that:
$$d(f(x),f(y)) \leq q d(x,y)$$
for every two elements ##x## and ##y## and for some number ##q \in [0,1)##.
Using the naturally induced norm, we have:
$$\lVert Q(f) - Q(g)\rVert^2 = \lVert f -g + \lambda(T(B(f-g)) \rVert^2 \stackrel{f-g=u}{=} \lVert u-\lambda T(B(u)) \rVert^2 = (u-\lambda T(B(u)),u-\lambda T(B(u))) = \lVert u\rVert^2 - 2\lambda(T(B(u)),u) + \lambda^2\lVert T(B(u))\rVert^2$$
At this point, we use the coercivity of ##\beta##, keeping in mind that we have:
$$\beta(f,h) = (T(B(f)),h) \wedge \beta(f,f) \geq C\lVert f\rVert^2$$
It follows that:
$$(T(B(u),u) = \beta(u,u) \geq C\lVert u\rVert^2$$
$$\lVert T(B(u))\rVert^2 \stackrel{Riesz}{=} \lVert B(u)\rVert^2 \leq M\lVert u\rVert^2$$
where ##M## is a positive number in the definition of boundedness of ##\beta##(since ##\beta## is continuous).
Substituting, we obtain the inequality:
$$\lVert Q(f) - Q(g)\rVert^2 \leq \lVert f-g\rVert^2 -2\lambda C\lVert f-g\rVert^2 +\lambda^2M^2\lVert f-g\rVert^2 = (1-2\lambda C + \lambda^2M^2) \lVert f-g\rVert^2$$
Now all that is left is to prove that we can pick ##\lambda##, such that the above coefficient ##(1-2\lambda C+\lambda^2M^2) \in (0,1)##. We keep in mind that trivially ##C\leq M##(because ##C\lVert u\rVert^2 \leq \beta(u,u) \leq M\lVert u\rVert^2## for any ##u## by definition).
We consider the following inequalities:
$$1 - 2\lambda C + \lambda^2M^2 <1 \Leftrightarrow 0<\lambda < \frac{2C}{M^2}$$
$$C<M \Rightarrow 1 - 2\lambda C + \lambda^2M^2> 0 $$
From this we see that the acceptable choice of ##\lambda##, would be ##0<\lambda<\frac{2C}{M^2}##. This value asserts that ##Q(f)## is a contraction, hence by Banach fixed point theorem, there is a unique fixed point of ##Q(f)##, which implies existence and uniqueness of ##f \in \mathcal{H}## such that for all ##h \in \mathcal{H}##:
$$F(h) = \beta(f,h)$$
as we have shown above when we defined ##Q##. This finishes our proof. Banach's theorem even gives algorithm of how we would arrive at this vector(as a limit of infinite power of ##Q## on an arbitrary vector ##f## from our Hilbert space, that is as: ##\lim_{n\rightarrow \infty} Q(Q(\dots Q(f)))## where ##Q## is applied in a composition ##n## times).
 
Last edited:
  • #102
12,879
9,512
I guess it's time to give both of you - @suremarc and @Antarres - the credit for solving the problem, since you both have had a correct strategy. Since this problem - Lemma of Babuška-Lax-Milgram - is a bit confusing, I'll add my proof. There was nothing wrong with either of yours, but for all other readers who might have had difficulties to follow:

If we define a continuous function ##B(f)(g):=\beta(f,g)## then Riesz' representation theorem gives us an isometric isomorphism ##T\, : \,\mathcal{H}^* \longrightarrow \mathcal{H}## such that for every ##B(f)\in \mathcal{H}^*## there is a unique ##T(B(f))## such that ##\|B(f)\|=\|T(B(f))\|## and
$$
B(f)(g)=\langle T(B(f)) ,g \rangle_\mathcal{H} = \beta(f,g)\quad \forall \,\, g\in \mathcal{H} \quad (*)
$$
or generally ##f^*(g)=\langle T(f^*),g \rangle_\mathcal{H}\,\, \forall \,\, g\in \mathcal{H} \quad (*)##

The functionals ##B(f)## are bounded since ##\beta## is continuous, i.e. ##\|B\|## is a finite real number. We get from our lower bound
\begin{align*}
C\|f\|^2&\leq |\beta(f,f)|=\langle T(B(f)) ,f \rangle_\mathcal{H}\\
&\leq \|T(B(f))\|\cdot\|f\|= \|B(f)\|\cdot\|f\| \leq \|B\|\cdot \|f\|^2
\end{align*}
hence ##0 < \dfrac{C}{\|B\|} \leq 1.## We now define the function
$$
Q(f):=f-k\cdot \left( T(B(f)) - T(F) \right)
$$
on ##\mathcal{H}## with a real number ##k\in \mathbb{R}-\{0\}.## A vector ##f^\dagger \in \mathcal{H}## is a fixed point of ##Q## iff ##T(B(f^\dagger)) - T(F)=0\,.## In general we have for all ##g\in \mathcal{H}##
\begin{align*}
T(B(f)) - T(F) \stackrel{(**)}{=} 0 &\Longleftrightarrow F(g)\stackrel{(**)}{=}B(f)(g) =\beta(f ,g)\stackrel{(*)}{=}\langle T(B(f)) ,g \rangle_\mathcal{H}\\
&\Longleftrightarrow F(g) \stackrel{(*)}{=} \langle T(F),g \rangle_\mathcal{H} \stackrel{(**)}{=} \langle T(B(f)) ,g \rangle_\mathcal{H}\\
&\Longleftrightarrow \langle T(B(f))-T(F),g\rangle_\mathcal{H} \stackrel{(**)}{=} 0
\end{align*}
again by Riesz' representation theorem and the equations above. As ##g\in \mathcal{H}## is arbitrary, we may set ##g:=T(B(f^\dagger ))-T(F)## for a fixed point of ##Q## and get ##\|T(B(f^\dagger ))-T(F)\|^2=0## hence
$$
B(f^\dagger)=\beta(f^\dagger,-)=F
$$
which has to be shown. Thus all what's left to show is, that such a unique fixed point ##f^\dagger## of ##Q## exists, which we will prove with Banach's fixed point theorem.
\begin{align*}
\|Q(f)-Q(g)\|^2 &= \| f-k(T(B(f))-T(F))-g+k(T(B(g))-T(F)) \|^2\\
&= \langle (f-g)-kT(B(f-g)),(f-g)-kT(B(f-g)) \rangle_\mathcal{H}\\
&\stackrel{(*)}{=}l, \|f-g\|^2 -2k\, \langle T(B(f-g)) , f-g\rangle_\mathcal{H} +k^2\,\|T(B(f-g))\|^2 \\
&\stackrel{(*)}{=} \|f-g\|^2 -2k \,\beta(f-g,f-g) +k^2\, \|B(f-g)\|^2 \\
&\leq \|f-g\|^2 -2k\, C\,\|f-g\|^2 + k^2\, \|B\|^2\,\|f-g\|^2 \\
&= \|f-g\|^2 \left( 1-2k\,C+k^2\,\|B\|^2 \right) \\
&\stackrel{\text{set }k:=C/\|B\|^2}{=} \|f-g\|^2 \left(1 - \dfrac{C^2}{\|B\|^2}\right)
\end{align*}
We have seen that ##\dfrac{C}{\|B\|} \in (0,1]## hence ##q:=1 - \dfrac{C^2}{\|B\|^2} \in [0,1)## and ##\|Q(f)-Q(g)\|^2=q\,\|f-g\|^2## and the statement follows from Banach's fixed point theorem.
 
Last edited:
  • Like
Likes Antarres and suremarc
  • #103
Hsopitalist
Gold Member
33
23
$$\begin{align*}
\lim_{n\to\infty}\cos\left(\frac{t}{\sqrt{n}}\right)^n&=\lim_{n\to\infty}e^{n\ln\left(\cos\left(\frac{t}{\sqrt{n}}\right)\right)}\\
&=\lim_{n\to\infty}e^{n\ln\left(1-\frac{t^2}{2n}+o(\frac{1}{n})\right)}\\
&=\lim_{n\to\infty}e^{n\left(-\frac{t^2}{2n}+o(\frac{1}{n})\right)}\\
&=e^{-\frac{t^2}{2}}
\end{align*}$$

Tried to PM the solver but couldn't. I have some questions about the steps here, particularly how the cosine was changed into the algebraic expression?
 
  • Like
Likes archaic
  • #104
508
126
  • #105
Hsopitalist
Gold Member
33
23
Oh man, thank you!
 
  • Like
Likes archaic
  • #106
508
126
  • #107
281
28
For #5, Let ##R## be a commutative ring. To say a subset ##I \subseteq R## is an ideal of ##R## means that ##I## is nonempty, ##I## is closed under addition and that for any ##r \in R## and ##i \in I##, we have ##ri \in I##. And to say ##\operatorname{char} R \neq 2## means ##2## is not the smallest positive integer ##n## such that ##n\cdot 1 = 0##. But i'm not sure how to relate this to matrix multiplication. Would it be possible to get another hint, please?
 
  • #108
12,879
9,512
For #5, Let ##R## be a commutative ring. To say a subset ##I \subseteq R## is an ideal of ##R## means that ##I## is nonempty, ##I## is closed under addition and that for any ##r \in R## and ##i \in I##, we have ##ri \in I##. And to say ##\operatorname{char} R \neq 2## means ##2## is not the smallest positive integer ##n## such that ##n\cdot 1 = 0##. But i'm not sure how to relate this to matrix multiplication. Would it be possible to get another hint, please?
The matrix has a certain additive decomposition (*) of which one part is in a certain ideal. The decomposition series of that ideal together with a certain property of (*) then solves the question.
 

Related Threads on Math Challenge - February 2020

Replies
86
Views
13K
Replies
150
Views
4K
Replies
77
Views
4K
Replies
64
Views
6K
Replies
51
Views
3K
Replies
18
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
20
Views
7K
Replies
40
Views
10K
Replies
7
Views
2K
Top