show that if w is continous on [0,1] and integral w*v on [0,1] equal zero then w(x)=0 for x in [0,1]
To prove that if the integral of w*v on [0,1] is equal to zero, then w(x) must be equal to zero for all x in [0,1], we can use a proof by contradiction.
Suppose there exists some x0 in [0,1] such that w(x0) is not equal to zero. Since w(x) is continuous on [0,1], we can use the intermediate value theorem. This theorem states that if f(x) is a continuous function on [a,b] and a value M lies between f(a) and f(b), then there exists a value c in [a,b] such that f(c) = M.
Since w(x) is not equal to zero at x0, we have w(x0) > 0 or w(x0) < 0. Without loss of generality, let's assume w(x0) > 0 (the proof is similar for w(x0) < 0).
Now, consider the function v(x) = sign(w(x0)). This function takes the value +1 if w(x0) > 0, and -1 if w(x0) < 0. It is piecewise continuous on [0,1] as sign(w(x0)) is a constant function.
Since w(x) is continuous on [0,1], the product w(x) * v(x) is also continuous on [0,1]. Now, integrating w(x) * v(x) over the interval [0,1], we get:
∫(0 to 1) w(x) * v(x) dx
Since v(x) is a constant function, we can take it out of the integral:
v(x) * ∫(0 to 1) w(x) dx
Now, since v(x) is a constant function, we can write the integral as:
v(x) * ∫(0 to 1) w(x) dx = v(x) * (W(1) - W(0))
where W(x) is the antiderivative of w(x).
Since v(x) is a constant function, we have:
v(x) * (W(1) - W(0)) = sign(w(x0)) * (W(1) - W(0))
But since we assumed that the integral of w(x) * v(x) over [0,1] is equal to zero, it follows that:
0 = sign(w(x0)) * (W(1) - W(0))
Since sign(w(x0)) is either +1 or -1, and (W(1) - W(0)) is a non-zero number (as w(x0) is not equal to zero), we cannot have 0 on the left side of the equation if the right side is non-zero. This contradicts our assumption that the integral of w(x) * v(x) over [0,1] is equal to zero.
Hence, our initial assumption that there exists an x0 in [0,1] such that w(x0) is not equal to zero must be false. Therefore, w(x) must be equal to zero for all x in [0,1].