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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 $f(1) >= 2$

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

2 Answers 2

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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. $\Box$

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  • $\begingroup$ Thanks this was helpful! $\endgroup$ Commented Dec 16, 2020 at 19:58
  • $\begingroup$ You're very welcome $\endgroup$
    – Mousedorff
    Commented Dec 16, 2020 at 19:59
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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<f(x)-f(1)<2-f(1)\\&\implies f(x)<2.\end{align}

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