show that if w is continous on [0,1] and integral wv on [0,1] equal zero then w(x)=0 for x in [0,1]
To prove that if the integral of the product of a continuous function w and a function v over the interval [0,1] is equal to zero, then w(x) = 0 for all x in [0,1], we can use the method of contradiction.
Let's assume that w(x) is not equal to 0 for some x in [0,1]. This means that there exists at least one point x0 in [0,1] where w(x0) is not equal to 0.
Since w is continuous on [0,1], we can apply the Intermediate Value Theorem. According to this theorem, if a function is continuous on an interval and takes on two distinct values at the endpoints, then it must also take on every value in between.
In this case, since w(x) is not equal to 0 for some x0 in [0,1], and we know that w(x) is continuous on [0,1], we can conclude that w(x) takes on every value between 0 and w(x0) in the interval [0,x0], and between w(x0) and 0 in the interval [x0,1].
Now, let's consider the integral of w(x)v(x) over the interval [0,1]:
∫[0,1] w(x)v(x) dx = ∫[0,x0] w(x)v(x) dx + ∫[x0,1] w(x)v(x) dx
Since w(x) takes on every value between 0 and w(x0) in the interval [0,x0], and between w(x0) and 0 in the interval [x0,1], we can conclude that w(x)v(x) is either positive, negative, or zero for each interval. Therefore, the integral of w(x)v(x) over each interval is non-zero, as it will have both positive and negative contributions.
This contradicts the assumption that the integral of w(x)v(x) over [0,1] is equal to zero. Hence, our initial assumption that w(x) is not equal to 0 for some x in [0,1] must be false.
Therefore, we can conclude that if the integral of w(x)v(x) over [0,1] is equal to zero, then w(x) = 0 for all x in [0,1].