- #101

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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).

$$F(x) = (T(F),x)$$

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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