Find $f(x_0)$ if $f(x) = 0$ for all $x \gt x_0$

Question

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function. Suppose $x_0 \in \mathbb{R}$ and $f(x) = 0$ for all $x \gt x_0$. Prove that $f(x_0) = 0$.

My try: Let $a_n = x_0 + \frac{1}{n}$. So we have $\displaystyle \lim_{n \to \infty} a_n = x_0 $ and $\displaystyle \lim_{n \to \infty} f(a_n)= 0$. By continuity we have $\displaystyle \lim_{n \to \infty} f(a_n) = f(\lim_{n \to \infty} a_n) = f(x_0)$. The result is $f(x_0) = 0$.

Is this solution correct? What are the different ways to prove that? Also I wonder if it's possible to prove $f(x_1) = 0$ for $x_1 \lt x_0$ by the same argument.

Answer 1

Your solution is correct. However your claim about $x_1$ is not. Consider:

$$f(x)=\begin{cases}x_0-x &\text { for } x < x_0 \\ 0&\text{ for } x\ge x_0\end{cases}$$

$f$ is continuous but $\ne 0$ for any $x < x_0$.

Answer 2

A topological way to see the result is to say that $$A = f^{-1}(\lbrace 0 \rbrace)$$

is closed (as the continuous preimage of $\lbrace 0 \rbrace$ which is closed), and contains $(x, +\infty)$, therefore it contains $x$.

Answer 3

Alternatively, if $f(x_o)$ was not zero, then there would be a neighbourhood of $x_o$ , say, $N(x_o, \delta)$ in which $f$ is non-zero by continuity of $f$.

But then there exist $x \gt x_o$ such that $f(x) \neq 0$ which contradict the hypothesis.!