Show that g is continuous on [0, 2], where we don’t have the function g(x)

Question

Suppose that $f : (0, 2] \to \mathbb{R}$ is a continuous function and that $\lim_{x \to 0}f(x) = L$ for some $L \in \mathbb{R}$. Define $g : [0, 2] \to \mathbb{R}$ via $g(0)=L$ and $g(x)= f(x)$ for $x \in (0,2]$.

(i) Show that $g$ is continuous on $[0, 2]$.

(ii) Show that $f$ is uniformly continuous on $(0, 2]$.

I'm struggling to know where to start! I understand that $|g(y)-g(x)|< \varepsilon$ for all $y \in [0,2]$ with $|y-x|<\delta$.

I just can't get the formal proof started.

Answer 1

Since $g(x)= f(x)$ on $(0, 1]$, and $f$ is continuous on $(0, 1]$, $g$ is continuous on $(0, 1]$. It is only necessary to show that $g$ is right-continuous at $0$. That is, that $\lim_{x\to 0^+} g(x)= L$ which is true because we were also told that $\lim_{x\to 0+} f(x)= L= g(0)$.

Saying that $f$ is "continuous" on $(0, 1]$ means "for every $x_0\in (0,1]$, given $\epsilon> 0$ there exist $\delta> 0$ such that if $|x- x_0|< \delta$ then $|f(x)- f(x_0)|<\epsilon$". Saying that $f$ is "uniformly continuous" means that, given $\epsilon$, we can choose a value of $\delta$ that will work for any $x_0$. There is a theorem that says that "if a function is continuous on a compact (i.e. closed and bounded) interval then it is uniformly continuous on that interval". So here you can choose some small $x_0$, assert that $f$ is uniformly continuous on $[x_0, 1]$, and then show that, given $\epsilon> 0$ there exist some smallest $\delta$ that will also work for $x\in (0, x_0)$.

Answer 2

Well, The definition of "$f$ is continuous at $x=a$" and the definition of "$\lim_{x\to a}f(x) = L$" are so similar that we can conclude: $f(x)$ is continuous at $a\iff \lim_{x\to a}f(x) = f(a)$.

Definiton: $\lim_{x\to a} f(x)=L$ if for every $\epsilon > 0$ we can find a $\delta >0$ so that whenever $0< |x-a|< \delta$ we will have $|f(x)- L| < \delta|$

Definition: $f$ in continuous at $x=a$ if for every $\epsilon > 0$ we can find a $\delta >0$ so that whenever $0< |x-a|< \delta$ we will have $|f(x)- f(a)| < \delta|$.

When talking $\lim_{x\to a}h(x)$ we are considering points where $x \ne a$, so when taking the limit of $\lim_{x\to 0}g(x)$ we are considering $x \ne 0$ so $g(x) = f(x)$ and $\lim_{x\to 0}g(x)= \lim_{x\to 0} f(x) = L = g(x)$. So $g$ is continuous at $x = 0$.

And for $a \ne 0$ then we can, in taking $\lim_{x\to a}g(x)$ consider just the $x$ within the range $(0, 2]; 0 < a \le 2$. So $f(x) = g(x)$ in that range and as $f$ ix continuous at $a$ we have $\lim_{x\to a} g(x)=\lim_{x\to a}f(x)=f(a)=g(a)$ so $g$ is continuous at $a$.

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If, for practice, you want a back to basics $\delta, \epsilon$ prove.

Let $\epsilon > 0$.

If $a\ne 0$ then $f(x)$ is continuous at $x=a$ so there is a $\delta$ so that $0<|x-a|< \delta$ implies $|f(x)-f(a)| < \epsilon$. But as $f(0)$ is undefined we can assum $|x-a|< \delta$ implies $x\in (0, 2]$ and so $f(x)=g(x)$ and $f(a)=g(a)$ so $|f(x)-f(a)|=|g(x)-g(a)|<\epsilon$ so $g(x)$ is continuous at $x = a\ne 0$.

And if $a = 0$ then $\lim_{x\to 0}f(x) = L$ so for every $\epsilon > 0$ there is a $\delta$ so that if $0< |x-0|< \delta$ then $|f(x) - L|< \epsilon$. But $f(x) = g(x)$ and $L = g(a)$ so $|g(x)-g(a)|< \epsilon$ so $g$ is continuous at $x=0$.