Prove if $f$ is continuous at a and $g$ is discontinuous at a, then $f + g$ is discontinuous at a

Question

Hello I want to prove: if $f$ is continuous at $a$ and $g$ is discontinuous at $a$, then $f + g$ is discontinuous at $a$. But with the $\epsilon - \delta$ definition of continuity and discontinuity (I see that the far more concise way of proving this would be to assume to the contrary that $f+g$ is continuous and then finding a contradiction from $g = (f + g) - f$).

So far I have:

Let $\epsilon > 0$ be given. Then since $f$ is continuous at $a$, there exists $\delta_{1} > 0$, such that $$|x - a| < \delta_{1} \implies |f(x) - f(a)| < \epsilon.$$

And since $g$ is discontinuous at $a$, there exists $\epsilon_{0} > 0$ such that for all $\delta_{2} > 0$, $$|x - a|<\delta_{2} \land |g(x) - g(a)| \ge \epsilon_{0}.$$

Now I want to show that there exists an $\epsilon_{1} > 0$ such that for any $\delta > 0$, $$|x - a| < \delta \land |f(x) + g(x) - (f(a) + g(a))| \ge \epsilon_{1}.$$

So now I need to find an appropriate $\epsilon_{1}$ and I think I should use the reverse triangle inequality to try to show this inequality holds $$0 < \epsilon_{1} \le ... \le |g(x) - g(a)| - |f(x) - f(a)| \le |g(x) - g(a) + f(x) - f(a)|$$

What seems problematic is the $|f(x) - f(a)|$ since there doesn't seem to be a way to bound it since $\delta$ can be any positive real number. I'm kind of wondering if this will even work or if another route needs to be taken. Any help would be much appreciated, thanks!

Answer 1

I'm not sure if your post is indicating that you know this, but there is a quick proof using a relatively well known fact: the sum of two continuous functions is continuous. Assume $f+g$ is continuous. Then since $f$ is continuous, $-f$ is continuous. Therefore $(f+g)+(-f)=g$ is continuous, contradiction.

If you don't know this fact already, I would recommend trying to prove it for yourself. The crucial step is to consider $\delta=\max{(\delta_f,\delta_g)}$. You can actually use the proof of the above statement to prove your question by contradiction. As you've noted, the fact that $\delta$ can be any real number can be problematic. However, you do have a statement that is true for all $\delta$, namely

$$\exists\epsilon_g\forall\delta_g\exists x_g\text{ such that }|x_g-a|<\delta_g\wedge |g(x_g)-g(a)|>\epsilon_g$$

This fact is the one that you want to leverage to get your result. I think by contradiction would be the best way to proceed. So, you assume $$(\forall\epsilon)(\exists\delta)(\forall x)(|x-a|<\delta\Rightarrow|f(x)+g(x)-f(a)+g(a)|<\epsilon)$$ and you have that $$(\forall\epsilon')(\exists\delta')(\forall x)(|x-a|<\delta'\Rightarrow|f(x)-f(a)|<\epsilon')$$ so therefore $$|g(x)-g(a)|=|f(x)+g(x)-(f(a)+g(a))-(f(x)-f(a))|\leq \epsilon+\epsilon'$$ by the triangle inequality, which contradicts that $g$ is discontinuous once you do a bit of tweaking of $\delta$'s and $\epsilon$'s.

Answer 2

You could use the sequential characterization of continuity, which says that $f$ is cts at $a$, if and only if for every sequence $x_n$ converging to $a$, we have that $f(x_n)$ converges to $f(a)$.

Now, let some arbitrary sequence converge to your point $a$.

We have that $(f+g)(x_n)$ = $f(x_n) + g(x_n)$, which does not converge to $f(a) + g(a)$ = $(f+g)(a)$, by assumption, since $g$ is known to be discts at $a$.

This shows that the function $f+g$ is discts at $a$, as required.

Answer 3

The direct approach will work. Remember that if $g$ is discontinuous at $a$, then there exists $\varepsilon_g>0$ such that for every $\delta>0$ there exists an $x$ within $\delta$ of $a$ such that $|g(x)-g(a)|\ge \varepsilon_g$. (Your solution left out the "there exists an $x$" part.)

Proof continues as follows. Since $f$ is continuous at $a$, there exists $\delta_f>0$ such that $|x-a|<\delta_f$ implies $|f(x)-f(a)|<\varepsilon_g/2$.

We'll show that $h:=f+g$ is discontinuous at $a$, with $\varepsilon_g/2$ as the magic value. So let $\delta>0$. Since $g$ is discontinuous at $a$, there exists $x$ such that $|x-a|<\min(\delta, \delta_f)$ and $|g(x)-g(a)|\ge\varepsilon_g$. For this $x$ we have $$|h(x)-h(a)|\stackrel{(1)}\ge|g(x)-g(a)|-|f(x)-f(a)|\stackrel{(2)}\ge\varepsilon_g-\varepsilon_g/2=\varepsilon_g/2.$$ Step (1) is the reverse triangle inequality, as you've done. In step (2) we use the fact that $|x-a|<\delta$ (so that $|g(x)-g(a)|\ge\varepsilon_g$), and the fact that $|x-a|<\delta_f$ (so that $|f(x)-f(a)|<\varepsilon/2$).