Prove a differentiable function between banach spaces is continuous

Question

I really don't know where to start with this problem. It intuitively makes sense from basic calculus knowledge but I have absolutely no clue how to show it. Definition of differentiable given by notes, question

Answer 1

According to a definition, in Wikipedia, $f(x)$ is differentiable at $x_0$, if there exists a linear operator $A(x_0)$, such that

$$\lim_{\|h\|\rightarrow0}\frac{\|f(x_0+h)-f(x_0)-A(x_0)h\|}{\|h\|}=0$$

Substitute $h$ by $x-x_0$ to get

$$\lim_{\|x-x_0\|\rightarrow0}\frac{\|f(x)-f(x_0)-A(x_0)h\|}{\|x-x_0\|}=0$$

It means that if $\|h\|=\|x-x_0\|$ is small enough, then

$$\|f(x)-f(x_0)-A(x_0)h\|<\|x-x_0\|$$

Also, from the properties of the norm

$$|\|a\|-\|b\||\leq\|a-b\|$$

Therefore

$$|\|f(x)-f(x_0)\|-\|A(x_0)h\||\leq \|f(x)-f(x_0)-A(x_0)h\|$$

As $A(x_0)$ is linear, it is bounded. Therefore, we have $\|A(x_0)h\|<M\|h\|$, for some limited $M$. As $\|h\|\rightarrow0$, so does $\|A(x_0)h\|$.

So, finally we get

$$\|f(x)-f(x_0)\| \leq \|x-x_0\|.$$

This is exactly what you need to have the definition of continuity, at $x_0$, completed.

Answer 2

Let $x$ be a point at which $f: U \subset E \to F$ is differentiable. For $\epsilon = 1$, there is $\delta > 0$ such that $\|h\| < \delta \implies \|f(x+h) - f(x) - Df(x)h\| < \|h\|$. Hence $\|f(x+h) - f(x)\| < \|Df(x) h\| + \epsilon \|h\|$ (using $\|a-b\| \ge \|a\| - \|b\|$). Thus $\|f(x+h) - f(x)\| < (1 + \|Df(x)\|)\|h\|$ for each $h$ with $\|h\| < \delta$ (recall that $Df(x)$ is continuous, so $\|Df(x)\| < \infty$). This shows that $\lim_{h \to 0} \|f(x+h) - f(x)\| = 0$, i.e. $\lim_{h \to 0} f(x+h) = f(x)$. So $f$ is continuous at $x$.