Question

(1) Suppose that f is continuous on la, b) except for at yi and y2. Prove that if eb for all continuous functions g : [a,b] → R, then f(y)メ0 and f(y)メ0.

Answer 1

Let
"x_0 elementof [a, b] \ {y_1, y_2} f(x_0) = 0 epsilon =
. We claim that .

Suppose this is not the case, and let $\epsilon=|f(x_0)|$ . Because
"x_0 elementof [a, b] \ {y_1, y_2} f(x_0) = 0 epsilon =
, continuity at this point implies there is some such that $(x_0-\delta,x_0+\delta)\subseteq [a,b]\setminus \{y_1,y_2\}$ or $(x_0-\delta,x_0]\subseteq [a,b]\setminus \{y_1,y_2\}$ or , and calling this neighborhood as , we see that for all $x\in I$ , we have

$|f(x)|>0.5\epsilon$

Let $g:[a,b]\rightarrow\mathbb R$ be defined as follows:

where one or both the limits has to be considered one-sided limit if the points happen to be . Note that the limits exist by continuity of .

Now, this function $g:[a,b]\rightarrow\mathbb R$ is continuous by definition and continuity of in $[a,b]\setminus \{y_1,y_2\}$ ; we also have

$\begin{align*}0&=\int_a^bf(x)g(x)dx\\ &=\int_a^b f(x)^2dx\\ &\geq \int_I|f(x)|^2dx\\ &>0.25\epsilon^2\int_Idx\\ &\geq 0.25\epsilon^2\delta\\ &>0\end{align*}$

which is absurd. Thus, our assumption that $f(x_0)\neq 0$ must be false. This proves the claim.

Now, we have shown that for all
"x_0 elementof [a, b] \ {y_1, y_2} f(x_0) = 0 epsilon =
. Since is not continuous at $\{y_1,y_2\}$ and continuous everywhere else in , we must have

$f(y_1)\neq\lim_{x\rightarrow y_1}f(x)=0,~~~~~f(y_2)\neq\lim_{x\rightarrow y_2}f(x)=0$