Exercise 2.2.10

Question

Let $f: \mathbb{R}^{2} \rightarrow \mathbb{R}$ be a continuous function. Show that for each $x \in \mathbb{R}$, the function $y \mapsto f(x, y)$ is continuous on $\mathbb{R}$, and for each $y \in \mathbb{R}$, the function $x \mapsto f(x, y)$ is continuous on $\mathbb{R}$.

Prakash Anand · Accepted Answer

We will prove only the first part since the second part is exactly the same.

Proof. Since $f$ is continuous, we know that for all $(x_{0}, y_{0}) \in ℝ^{2}$ , given $𝜖 > 0 \exists δ > 0$ such that if $d_{(ℝ^{2}, l^{2})} ((x, y), (x_{0}, y_{0})) < δ$ , then $d_{ℝ} (f (x, y), f (x_{0}, y_{0})) < 𝜖$ .

Let $x_{0} \in ℝ$ and given $𝜖^{'} > 0$ , take $δ^{'} = δ > 0$ .

If $d_{ℝ} (y, y_{0}) < δ^{'}$ we get that

| y - y_{0} | < δ^{'} \Rightarrow {(y - y_{0})}^{2} < {(δ^{'})}^{2} .

Also note that

| x_{0} - x_{0} | = 0 \Rightarrow {(x_{0} - x_{0})}^{2} = 0

So

{(x_{0} - x_{0})}^{2} + {(y - x_{0})}^{2} < {(δ^{'})}^{2}

Thus,

\sqrt{{(x_{0} - x_{0})}^{2} + {(y - x_{0})}^{2}} < δ^{'} = δ

But by definition this is

d_{(ℝ^{2}, l^{2})} ((x_{0}, y), (x_{0}, y_{0})) < δ

And since $f$ is continuous we have

d_{(ℝ, l^{2})} (f (x_{0}, y), f (x_{0}, y_{0})) < 𝜖^{'}

Since $x_{0}$ was chosen arbitrarily we know that for each $x \in ℝ$ , the function $y \mapsto f (x, y)$ is continuous on $ℝ$ . □

mathchakra

2021-12-19 18:19

Exercise 2.2.10

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Exercise 2.2.10

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