# Assume $f:[0,1] \rightarrow \mathbb{R}$ is continuous and that $f(x) \geq 2 \quad \text{ if } 0 \leq x \lt 1$. Show that $f(1) \geq 2$

This question comes from Advanced Calculus, Fitzpatrick. Section 3.1 exercise 7

Suppose that the function $$f:[0,1] \rightarrow \mathbb{R}$$ is continuous and that $$f(x) \geq 2 \quad \text{ if } 0 \leq x \lt 1$$ Show that $$f(1) \geq 2$$

Fitzpatrick uses the sequence definition of continuity, there is a similar question [here] (Suppose that the function $f:[0,1]\rightarrow \mathbb{R}$ is continuous and that $f\left(x\right)>2$)

I wanted to do this using epsilon-delta method. Is the following valid? Is there a simpler way to write this proof?

My attempt

First, assume $$f(1) < 2$$. Then there exists some $$\varepsilon > 0$$, call it $$\varepsilon_0$$, such that $$2 - f(1) > \varepsilon_0$$. Show this creates a contradiction.

Let $$x_n = 1 - 1/n$$ for all $$\mathbb{N}$$. Then $$x_n$$ is a sequence in $$[0,1]$$ that converges to 1. By definition, this is

$$\forall \delta_1 > 0 \; \exists N \in \mathbb{N}, \forall n \geq N \quad \vert x_n - 1 \vert < \delta_1$$

Since $$f$$ is continuous on $$[0,1]$$ it is continuous at $$1$$, therefore we have

$$\forall \varepsilon > 0 \; \exists \delta > 0 \; \forall x \in [0,1]\quad \vert x - 1 \vert < \delta \rightarrow \vert f(x) - f(1) \vert < \varepsilon$$

If we let $$\varepsilon = \varepsilon_0$$ and $$\delta = \delta_1$$. Then we know $$\vert x_n - 1 \vert < \delta_1$$, therefore we can conclude $$\vert f(x_n) - f(1) \vert < \varepsilon_0$$.

Furthermore, we know $$\{f(x_n)\} >= 2$$ for all $$n$$, so

$$\vert 2 - f(1) \vert \leq \vert f(x_n) - f(1) \vert < \varepsilon_0$$ $$\vert 2 - f(1) \vert < \varepsilon_0$$ $$-\varepsilon_0 < 2 - f(1) < \varepsilon_0$$

However this contradicts our assumption that $$2 - f(1) > \varepsilon_0$$

Therefore $$f(1) >= 2$$

• Mentioning the sequence is redundant, but the idea and rest of the reasoning is good. Dec 16, 2020 at 19:40
• @NL1992 Nope when he took $\delta = \delta_1$ that was a mistake, you can't take any value you want for $\delta$ Dec 16, 2020 at 19:41
• @Laassila souhayl you are correct. $\delta$ is given to you since you chose $\epsilon_0$ Dec 16, 2020 at 19:45

You haven't exactly done this using the $$\epsilon-\delta$$ definition, though. Someone else has already commented on your proof so I'll just present my own argument.

Suppose that $$f(1) < 2$$. Since $$f$$ is continuous on $$[0,1]$$:

$$\lim_{x \to 1^-} f(x) = f(1) < 2$$

So, for each $$\epsilon > 0$$, there is a $$\delta > 0$$ such that:

$$1-x < \delta \implies |f(x)-f(1)| < \epsilon$$

Let $$\epsilon = 2-f(1)$$. Then, for a sufficiently small $$\delta > 0$$, we have:

$$1-x < \delta \implies f(1)-\epsilon < f(x) < f(1)+\epsilon$$

which implies that $$f(x) < 2$$ whenever $$x \in (1-\delta,1]$$. That's a contradiction. $$\Box$$

• Thanks this was helpful! Dec 16, 2020 at 19:58
• You're very welcome Dec 16, 2020 at 19:59

Yout approach is correct, but you don't need to use sequences. Suppose that $$f(1)<2$$. If $$f(1)<2$$, then there is some $$\delta>0$$ such that$$|x-1|<\delta\implies\bigl|f(x)-f(1)\bigr|<2-f(1).$$But\begin{align}\bigl|f(x)-f(1)\bigr|<2-f(1)&\iff f(1)-2