Suppose we consider the vector space of real-valued continuous functions $V:= C[-1,1]$. Then $\phi(f):= f(0) \; \forall \; f \in V$ is a linear functional.
I want to show that $\phi(f)$ cannot be represented as $\int_{-1}^{1} f(x) g(x) dx$ for any $g \in V$.
How would I go about showing this ? I would appreciate some hints, without invoking advanced concepts.
EDIT :This problem appears in an exercise on linear algebra, to show that the Riesz Representation Theorem on finite dimensional spaces may not hold on function spaces without additional assumptions. Hence I assume there must be a simple way to show this.
Here is one way, suppose $\phi(f) = \int f \cdot g$.
Let $r_n$ be the function given by straight line interpolation between the points $(-1,1), (-{1 \over n},1), (0,0), ({1 \over n}, 1), (1,1))$.
Let $f_n = g r_n$, then $\phi(f_n) = 0 \ge \int_{[-1,-{1 \over n}] \cup [{1 \over n}, 1]} g^2$, and since $g$ is continuous, we see that $g = 0$, which gives $\phi = 0$, a contradiction.
Here is another way:
Let $|g(x)| \le B$, then if $\phi(f) = \int f \cdot g$, we have $\phi(f) \le B \int |f|$.
Now let $f_n = \max(0, 1-n|x|)$, then we have $\phi(f_n) = 1 \le B {1 \over n}$ for all $n$, which is a contradiction for large $n$.