# Prob. 12, Sec. 5.2, in Bartle & Sherbert's INTRO TO REAL ANALYSIS: Continuity of additive functions

Here is Prob. 12, Sec. 5.2, in the book Introduction To Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:

A function $f \colon \mathbb{R} \to \mathbb{R}$ is said to be additive if $f(x+y)= f(x) + f(y)$ for all $x, y$ in $\mathbb{R}$. Prove that if $f$ is continuous at some point $x_0$, then it is continuous at every point of $\mathbb{R}$.

My Attempt:

Let $c$ be an arbitrary real number.

First suppose that $x_0 = 0$. As $f$ is continuous at the point $0$, so given any real number $\varepsilon > 0$ we can find a real number $\delta > 0$ such that $$\lvert f(x) \rvert = \lvert f(x) - 0 \rvert = \lvert f(x) - f(0) \rvert < \varepsilon$$ for all $x \in \mathbb{R}$ for which $$\lvert x \rvert = \lvert x-0 \rvert < \delta.$$ Then $$\lvert f(x) - f(c) \rvert = \lvert f( x - c ) \rvert < \varepsilon$$ for all $x \in \mathbb{R}$ such that $$\lvert x-c \rvert < \delta.$$ Thus it follows that $f$ is continuous at every point $c \in \mathbb{R}$.

Am I right?

Now let us suppose that $x_0 \neq 0$. As $f$ is continuous at $x_0$, so, for every real number $\varepsilon > 0$ we can find a real number $\delta > 0$ such that $$\left\lvert f \left(x - x_0 \right) \right\rvert = \left\lvert f(x) - f \left( x_0 \right) \right\rvert < \varepsilon$$ for all real numbers $x$ which satisfy $$\left\lvert x- x_0 \right\rvert < \delta.$$

Now if we could show from here that $f$ is continuous at $0$, then as before we will have managed to show that $f$ is continuous at every real number $c$.

What next? How to proceed from here to show that our function $f$ is continuous at $0$? Or, is there any other route?

• Would you agree that a systematic call to check your work is associated much more closely with "seeking personal advice" rather than just asking mathematical questions? – uniquesolution Aug 14 '18 at 12:20
• @uniquesolution I'm sorry but do you disapprove of my asking questions in the manner I do? What if I arrive at a wrong conclusion during the steps of my proof? Hence my asking "Am I right?"! – Saaqib Mahmood Aug 14 '18 at 13:05
• I appreciate your keen efforts to study mathematics. I do think however that it takes some non-trivial presumptuousness to bombard this site with requests to practically tutor your self-study. I think that this site is not the place for that, and that's why I think such posts as yours ought to be closed on the grounds that they seek personal advice. – uniquesolution Aug 14 '18 at 14:08
• @uniquesolution asking to check one’s work does not come under “seeking personal advice”. There is even a tag called proof-verification for the same. – Brahadeesh Aug 14 '18 at 19:42
• @uniquesolution that is pretty clearly not the intended use of the seeking personal advice close button. – spaceisdarkgreen Aug 14 '18 at 19:44

You are almost there. Let $| t-0 | = |t| < \delta$, where $\delta$ is the one you have in the second box. Then $t + x_0 = x\in \mathbb{R}$, so $|t| = |x - x_0| < \delta$.
Thus, $|f(t) - f(0)| = |f(t)| = |f(x-x_0)| = |f(x) - f(x_0)| < \epsilon$ for $|t| < \delta$.
So $f$ is continuous at $0$.