Why continuity follows for $|f(a+h)-f(a)| \leq k|h|$, where $k$ is a scalar constant

Question

The other day I asked a question related to the problem below, then I realized that I wasn't sure how the inequality $|f(a+h)-f(a)| \leq C(|h_1|+|h_2)|$--very last line of their proof--implied continuity of $f$. Someone then commented that we can extend the inequality by writing $\lvert f(a+h) - f(a) \rvert \leq C (\lvert h_1 \rvert + \lvert h_2 \rvert) \leq C( \lvert h \rvert + \lvert h \rvert) \leq 2 C \lvert h \rvert$ but I still didn't see how continuity followed

My thinking was to think in terms of the delta-epsillon definition but then I was stuck trying to show that $|f(a+h)-f(a)| \leq |h|$, which a commenter pointed out that it failed in some cases, such as the case of $f(x)=3x$.

My Question

Could you explain why this inequality in their last line implies continuity and why scalar constants on $|h|$ still allow for continuity to follow?

Thanks, and here's the link to where I asked the other question. In the comments you'll see where we discuss continuity.

Proving continuity when partial derivatives are bounded on open set

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Solution

Answer 1

Let $\varepsilon > 0$. Then for $\|h\| < \min\left\{\frac{\varepsilon}{ C\sqrt2}, r\right\}$ we have

$$|f(a+h) - f(a)| \le C(|h_1| + |h_2|) \stackrel{CSB}{\le} C\sqrt{2}\sqrt{|h_1|^2+|h_2|^2} = C\sqrt2 \|h\| < \varepsilon$$

Answer 2

Basically, the proof says that for any $r>0$ and $\lvert\textbf{h}\rvert < r$ the difference $\lvert f(a+h)-f(a) \rvert$ is less than $C(|h_1|+|h_2|)$.

Now we know, that we can make $r$ as small as we want. Since this holds for any $r>0$ (that is small enough), the only possible thing this can me is that $|f(a+h)-f(a)| \to 0$ as $r \to 0$.

That is for any small distance $r>0$, the difference between function values that are less than $r$ distant from $a$ and $f(a)$ is lesser than $C(|h_1|+|h_2|)$. This should intuitively represent the concept of continuity.

To convert this idea into $\varepsilon$-$\delta$ proof, you just make your $\varepsilon:=C(|h_1|+|h_2|)$ and find $r$ (as your $\delta$) accordingly.