- #76

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I will assume here that ##\beta(x,x) = 0## is equivalent to ##x=0##, since that wasn't mentioned, but if we take that all vectors have non-zero norm, we exclude the zero vector(cause the form is bilinear) from the space, which shouldn't happen.

Denote ##K = \ker F## which is a subset of ##\mathcal{H}##(a closed subset, by continuity of ##F##), we assert that ##K^\perp## is one-dimensional(if we exclude the trivial case where kernel would take up the whole space, for which choice ##f^\dagger=0## would suffice). Suppose that it isn't, that there are two independent non-zero vectors ##v_1## and ##v_2## in ##\mathcal{H}##, that belong to ##K^\perp##. Then we have that ##F(v_1)v_2 - F(v_2)v_1## is a non-zero vector from ##K^\perp##, but by linearity of ##F## we have that

$$F(F(v_1)v_2-F(v_2)v_1) = 0$$

which is a contradiction as it would indicate ##F(v_1)v_2 - F(v_2)v_1 \in K##. We can extend this argument to any number of dimensions.

Assume ##K^\perp## is spanned by ##n## vectors. Then we create sum of form:

$$v=\sum_i a_i F(y_1)F(y_2)\dots y_i F(y_{i+1})\dots F(y_n) $$

such that ##\sum_i a_i=0##.

Then ##F(v)=0##, similar to above so its a contradiction.

So ##K^\perp## is one-dimensional.

Let's find a vector ##w\in K^\perp## such that ##F(w) = \beta(w,w)##.

Choose:

$$w= \frac{yF(y)}{\beta(y,y)}$$

for some non-zero ##y\in K^\perp##. By substitution we see that it satisfies the relation above that we wanted to impose. We can span ##K^\perp## with ##w##.

That means that any vector from ##\mathcal{H}## can be decomposed in the form ##v = u + kw##, where ##k## is a scalar, ##u\in K## and ##w \in K^\perp##. It follows that:

$$F(x) = kF(w) = k\beta(w,w) = \beta(w,u + kw) = \beta(w,x)$$

where we used ##\beta(u,w)=0## - the orthogonal complement definition, and ##F(u) = 0## cause ##u\in K##. Therefore we constructed ##w\in\mathcal{H}## such that ##F(x) = \beta(w,x)##, so in our formula ##w=f^\dagger##.

Let's prove uniqueness. Suppose there are ##f_1^\dagger## and ##f_2^\dagger## that both satisfy the identity ##F(x) = \beta(f,x)## for all ##x\in\mathcal{H}##.

Then, take ##x = f_1^\dagger - f_2^\dagger##. It follows that:

$$F(x) = \beta(f_1^\dagger, f_1^\dagger-f_2^\dagger)$$

but also:

$$F(x) = \beta(f_2^\dagger,f_1^\dagger-f_2^\dagger)$$

Subtracting the identities above we find:

$$\beta(f_1^\dagger - f_2^\dagger, f_1^\dagger-f_2^\dagger) = 0 \Rightarrow f_1^\dagger = f_2^\dagger.$$

The uniqueness is thus proved.

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