# Math Challenge - February 2020

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@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).

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Mentor
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.

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Antarres and suremarc
Hsopitalist
Gold Member
\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?

archaic
Hsopitalist
Gold Member
Oh man, thank you!

archaic
Oh man, thank you!
You are welcome!

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?

Mentor
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.