# Why exactly is the sum/difference of two (and possibly more) differentiable and continous functions also inherit this property?

OKay right off the bat, this seems like an intuitive and self-explantory result, at least when I started doing calculus it would seem self-explanatory. But now I have reached the point of math where I seriously just can't be sure of anything anymore.

To specify, I guess the function should be differentiable on an open interval in $\mathbb{R}$ and continuous on any interval or if that isn't strong enough, make it continuous on an closed interval.

If the cases are different, then show them all to me

-
These are all limit results. For continuity or differentiability and a sum of say two functions, it is the result $\lim_{x\to a}(f(x)+g(x))=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$, if the limits on the right exist. The proof is straightforward $\epsilon$-$\delta$. For product, the details are somewhat more messy. –  André Nicolas Dec 25 '12 at 20:39
I'm fairly certain that's because $+$ is continuous. –  Karolis Juodelė Dec 25 '12 at 20:42

Take $f(x)$ and $g(x)$ and assume, first, that they are both continuous. Then using the fact that each of $f(x)$ and $g(x)$ is continuous (strictly defined), then establish that $h(x) = f(x) + g(x)$ (or $h(x) = f(x) - g(x)$) must also be continuous.

Put differently: What is required for $h(x)$ to be continuous? Establish that we can satisfy this requirement (definition) provided $f(x)$ is continuous, by definition, and likewise $g(x)$ is continuous.

Do the same with $f(x)$ and $g(x)$ being differentiable functions. What can we then conclude about the differentiability of $h(x) = f(x) \pm g(x)$?

You'll see that your intuition can readily be justified.

• Recall: If $\lim_{x\to a}f(x)$ and $\lim_{x\to a}g(x)$ both exist, then if $h(x) = f(x) + g(x)$, we have that $\lim_{x\to a}h(x)$ exists and $$\lim_{x\to a}h(x) = \lim_{x\to a}(f(x) + g(x)) = \lim_{x\to a}f(x) + \lim_{x\to a}g(x).$$
• The above can be proven using the standard $\epsilon-\delta$ definition.

On a side note: I don't know of any serious math student who *doesn'*t, at some point along the way, question the certainty of what one s\he learned. That's to be expected, and once you get on firm ground again, which you will, you'll likely encounter that feeling again, down the road. It's actually a good sign that you are learning, and are in the process of acquiring an understanding of what you've previously learned at a far deeper level than ever before...

-

Hint: if you write out the definitions of continuity and differentiability of the sum of two functions, it should become very clear to you. At it's core, the answer is that because the limit of a sum is (equal to) the sum of the limits.

-

Look at the following inclusions: $$i_1:\mathbb{R}\rightarrow\mathbb{R}^2:x\mapsto (x,0),$$ $$i_2:\mathbb{R}\rightarrow\mathbb{R}^2:x\mapsto (0,x).$$ Note that $i_1$ and $i_2$ are continuous and differentiable functions.

Now observe that the "sum" function: $$+:\mathbb{R}^2\rightarrow\mathbb{R}:(x,y)\mapsto x+y$$ is also a continuous and differentiable mapping.

If you have two functions $f,g:\mathbb{R}\rightarrow\mathbb{R}$ that are continuous (or differentiable), then their product function: $$f\times g=(f,g):\mathbb{R}\rightarrow\mathbb{R}^2:x\mapsto (f(x),g(x))$$ also is continous (or differentiable).

Now notice that $$f+g=+\circ (f\times g).$$ Since (fortunately!) continuity (and differentiability) is preserved by the operation of composition of functions, the sum $f+g$ is also continuous (or differentiable).

IMHO that is the primary reason why the property you are interested in holds.

-