Bounded linear function implication

Suppose $f$ is linear, and $f(x) \geq B$ for all $x$. Choose any $x_0$. Then $f(\lambda x_0) = \lambda f(x_0) \geq B$ for all $\lambda$.

Take $\lambda >0$, which gives $f(x_0) \geq \frac{B}{\lambda}$, and since $\lambda$ is arbitrary, we have $f(x_0) \geq 0$.

Now take $\lambda <0$, which gives $f(x_0) \leq \frac{B}{\lambda}$, which now results in $f(x_0) \leq 0$.

Together, this gives $f(x_0) = 0$. Since $x_0$ was arbitrary, we have $f=0$.

Following the above, $f(x_0) \geq \frac{B}{\lambda}$ and $f(x_0) \geq B$.

For $\lambda > 0$. If $f(x_0) < 0$, then $\frac{B}{\lambda}$ could be greater than $f(x_0)$ for some $\lambda$, which is a contradiction. Hence, $f(x_0) \geq 0$.

For $\lambda \leq 0$. If $f(x_0) \leq 0$, then $\lambda f(x_0) \geq 0 \geq B$ holds. Hence, $f(x_0) \leq 0$.

As desired, $f(x_0) = 0$ for any $x_0$.