# Math Challenge - February 2020

• Challenge
• Featured
@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.