Representative of Riesz problem

[pgandalf]

Homework Statement

Find the representative of riesz of the following linear transformation:
$$\mathbb{T:}P_1\rightarrow{R}$$ defined by $$\mathbb{T}\left ( p \right )\mathbb{=}$$$$p\left ( \alpha \right )$$ where $$\alpha$$ is a fixed real number
(Considering in $$P_1$$ the inner product: $$\left<{p,q}\right>$$$$=\displaystyle\int_{0}^{1} p(t)q(t)\, dt$$ )

Homework Equations

formula of orthogonal projection $$P_s(w)=\left<{w,v}\right>_v$$
norm induced by the inner product=inner product of a vector with itself
Representative of Riesz (q): $$T(p)=\left<{p,q}\right>$$
Theorem of riesz: $$q=\bar{T}(e_{1})(e_{1})+...+\bar{T}(e_{n})(e_{n})$$ being $$\left\{{e_{1},...,e_{n}}\right\}$$ an orthonormal base of the vectorial space.

The Attempt at a Solution

All right what I did was find an orthonormal base of the polynomials of grade less or equal to 1.
For that I started from the base $$\left\{{1,t}\right\}$$ and I orthonormalized it finding the orthogonal projection of the vector $$t$$ in the space generated by the $$1$$ (which I call s from now) and then I find $$u/u=P_s(t)$$ and u is orthogonal to 1.

Ok,using the formula of orthogonal projection $$P_s(w)=\left<{w,v}\right>_v$$ I find the orthogonal projection using the inner product of the problem and it results 1/2 so $$u=t-\displaystyle\frac{1}{2}$$ and I normalize the base using the norm induced by the inner product and the norm of 1 is 1 and the norm of t is $$\displaystyle\frac{1}{\sqrt{12}}$$ so $$\left\{{1,\displaystyle\frac{t-\frac{1}{2}}{\sqrt{12}}}\right\}$$ is an orthonormal base of $$P_1$$ and I transform this base using the end of the theorem of riesz
and it results $$q=\bar{T}(1)(1)+\bar{T}\left(\displaystyle\frac{t-\frac{1}{2}}{\sqrt{12}}}\right)\left(\displaystyle\frac{t-\frac{1}{2}}{\sqrt{12}}}\right)$$

Here is my doubt: the transformed of 1 is 1? because if the transformed of any polynomial results the polynomial evaluated in alpha it results 1. The problem is that it doesn't verify because for instance for the polynomial 7t according to Riesz the transformed should be $$7\alpha$$ and that be equal to the inner product between 7t and the representative of Riesz (q).

With the q that I found : $$q=1+\left(\displaystyle\frac{\alpha-\frac{1}{2}}{\sqrt{12}}\right)\left(\displaystyle\frac{t-\frac{1}{2}}{\sqrt{12}}\right)$$ It doesn't verify (It should result 7$$\alpha$$ and it results a fraction of difference and the terms with alpha don't result 7 ?

Oh the formula that I'm using to verify is the one of the representative of riesz:
$$\mathbb{T}(p)=\left<{p,q}\right>$$
so if $$p(t)=7t$$,$$\left<{p,q}\right>=7\alpha$$ and this doesn't verify.

Sorry for the extent but in this way you can see what I'm missing or if you have an easier way of solving it you can tell me as well ;) Also although I have studied english for 8 years and passed the FCE I don't know about the language in maths although it's quite universal. If there's something that you don't understand ask me

 

[morphism]

I think you might be over-complicating things. Let's start by assuming that q(t)=a+bt for some real numbers a and b. We want the following equation to hold for all p in P_1:

$$\mathbb{T}p = \int_0^1 p(t)q(t) dt.$$

Now try a couple of simple p's and then solve for a and b.

 

[pgandalf]

You were right! I took $$p(t)=t+1$$ and $$p(t)=t+2$$ and I made the inner product of each of them with $$q(t)=a+b(t)$$ and then I used $$T(p)=p(\alpha)$$ (hyphotesis of the problem) and I equalled it to $$<p,q>$$ (because of the formula of the representative of riesz, $$T(p)=<p,q>$$ ) and I solved the system and I obtained $$q(t)=-6\alpha+4+(12\alpha-6)t$$ and then I used the polynomial $$p(t)=7t$$ to verify and indeed it does verify. Thank you a lot again!