How to convince myself that I can take n=1 here?

[Hall]

Homework Statement:

In the linear space $P_n$ of all real polynomials of degree $\leq$ $n$, define
$\langle f,g \rangle = \sum_{k=0}^{n} f \left(\frac{k}{n}\right) ~g\left( \frac{k}{n} \right)$
If $f(t)=t$, find all real polynomials $g$ orthogonal to $f$.

Relevant Equations:

$\langle f(t)=t, g \rangle = \sum_{k=0}^{n} \frac{k}{n} ~g\left( \frac{k}{n} \right)$

The Homework Statement reads the question.

We have
$$
\langle f,g \rangle = \sum_{k=0}^{n} f\left(\frac{k}{n}\right) ~g\left( \frac{k}{n} \right)
$$
If $f(t) = t$, we have degree of $f$ is $1$, so, should I take $n = 1$ in the above inner product formula and proceed as follows
$$
\langle f(t)=t, g \rangle = \sum_{k=0}^{1} \frac{k}{1} ~g\left( \frac{k}{1} \right)$$
$$\sum_{k=0}{1} \frac{k}{1} ~g\left( \frac{k}{1} \right) = 0 $$
$$f(0) ~g(0) + f(1) ~g(1) = 0$$
$$g(1) = 0 ~~~\implies ~~ g(t) = a(1-t) ~~~~~~~~~~~~\text{for any real a}
?
$$
I'm finding it hard to convince myself that I can take $n=1$ just because the degree of $f$ is 1, the $n$ given to us is the fixed degree upto which polynomials are allowed in our linear space $P_n$.

If I don't take $n=1$ the thing would get a little messier:
$$\sum_{k=0}^{n} \frac{k}{n} g\left( \frac{k}{n} \right) = 0 $$
$$\frac{1}{n} \sum_{k=1}^{n} k ~g\left(\frac{k}{n} \right) = 0 $$
$$\sum_{k=1}^{n} k ~g\left(\frac{k}{n} \right) = 0 $$
$$\text{One possible solution for g is} $$
$$g (t) = a~(t - 1/n) (t- 2/n) \cdots (t - n/n)
$$

Do I have to take $n=1$ or not?

 

[Mark44]

I'm finding it hard to convince myself that I can take n=1 just because the degree of f is 1, the n given to us is the fixed degree upto which polynomials are allowed in our linear space P_n.

I agree. It could be that n = 3, for instance, but the function f(t) = t would still belong to that space. I don't think you can assume that n = 1.