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

$|\hspace{0.1cm}||a||-||b||\hspace{0.1cm}|\leq||a-b||$

Therefore

$|\hspace{0.1cm}||f(x)-f(x_0)||-||A(x_0)h||\hspace{0.1cm}|\leq ||f(x)-f(x_0)-A(x_0)h||$

As $A(x_0)$ is linear, it is bounded. Therefore, we have $||A(x_0)h||

So, finally we get

$\hspace{0.1cm}||f(x)-f(x_0)||\hspace{0.1cm} \leq ||x-x_0||$

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 $x$.