Provided $f$ is continuous at $x_0$, and $f(x+y) = f(x) + f(y)$ prove $f$ is continuous everywhere.

Question

My attempt...

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By definition, whenever $|x- x_0| < \delta$ we have $|f(x) - f(x_0)| < \epsilon$. Observing that

\begin{align} |f(x) - f(y)| &= |f(x -x_0 + x_0) + f(y)| = |f(x-x_0) + f(x_0) - f(y)| \newline \newline &\leq \epsilon + |f(y) - f(x_0)| = \epsilon + |f(y-x_0)|... \end{align}

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Here I need to choose a delta that can depends on $\epsilon$ and $y$ s.t. whenever $0<|x-y|< \delta$ then the above inequality is bounded by any $\epsilon$.

I'm also, in general, having trouble understanding this concept of continuity on an interval. I believe the structure of the definition is: for any $\epsilon> 0$ and any number $y$ in the interval, there exists a $\delta$ that depends on $\epsilon$ and $y$ such that for all $x$ in the interval and $|x - y | < \delta$ then $|f(x) - f(y)| < \epsilon$.

This definition makes me tempted to just choose y to be in the same delta neighborhood as $x$ in the given statement, but that constricts continuity to a small interval.

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Edit: This question assumes no knowledge of Lebesgue measure theory.

Answer 1

Sketch/Scratch work:

• First, prove $f(0)=0$ since $f(0+0)=f(0)+f(0)$.

• Second, prove $f(-y)=-f(y)$ since $f(y-y)=f(y)+f(-y)$.

• Suppose that $|x-y|<\delta$. Then $$ |f(x)-f(y)|=|f(x)-f(y)+f(x_0)-f(x_0)|=|f((x-y)+x_0)-f(x_0)|. $$

• Observe that since $|x-y|<\delta$, you have a version of the continuity statement for $x_0$ (replace $x-y$ by $w$ where $|w|<\delta$ if you don't see it).

More details:

Let $\varepsilon>0$, since $f$ is continuous at $x_0$, there exists a $\delta$ so that if $|x-x_0|<\delta$, then $|f(x)-f(x_0)|<\varepsilon$. Fix $x_1$ and let $y$ be such that $|y-x_1|<\delta$. If you can prove that $|f(y)-f(x_1)|<\varepsilon$, then you have proved continuity at $x_1$. Since the choice of $x_1$ is arbitrary, $f$ is continuous.

Answer 2

Let $a\in\mathbb R.$ If the function $f(x)$ is continuous at $x_0,$ then the function $g(x):=f(x+a)=f(x)+f(a)$ is continuous at $x_0-a,$ and so is $f(x)=g(x)-f(a).$ Since $a$ is an arbitrary real number, so is $x_0-a;$ thus $f(x)$ is continuous everywhere.

Answer 3

For any $\epsilon>0$ there is a $\delta>0$ such that if $|x-x_0| < \delta$ then $|f(x)-f(x_0)| < \epsilon$.

Since $f(x)-f(x_0) = f(x-x_0)$ we see that this can be written as for any $\epsilon>0$ there is a $\delta>0$ such that if $|h| < \delta$ then $|f(h)| < \epsilon$.

Now pick some other $x_1$ and note that since $f(x) -f(x_1) = f(x-x_1)$, then if $|x-x_1| < \delta$ we must have $|f(x)-f(x_1)| < \epsilon$.