I don't know what a round ball is. I hope this is just an unnecessary Detail but if this is important to solve this excercise please let me know.
The original text is in German:
Offene und abgeschlossene runde Bälle in $\mathbb{R}^n$ sind konvex.
My thoughts are:
A subset $X$ in $\mathbb{R}^n$ is convex if and only if I take two arbitrary Points $a,b\in X$ then $\forall t\in[0,1]$ $a+t(b-a)=x_t\in X$.
I also know what a closed and open ball is. $B(x, r) = \{y ∈ X : d(x, y) < r\}$
where $X$ is $\mathbb{R}^n$ and $d(x,y)=\sqrt{(x_1-y_1)^2+...+(x_n-y_n)^2}$
This is an open ball, the closed ball would then have to be with $\leq$
I take $a,b\in B(x,r)$ then $d(x,a)<r$ and $d(x,b)<r$ I now have to show $d(x,x_t)<r\iff \sqrt{(x_1-(a_1+t(b_1-a_1)))^2+...+(x_n-(a_n+t(b_n-a_n)))^2}<r$
My idea was to loke at the equation componentwise but I don't know how to continue:
I look at one summand under the root sign, it has the form
$(x_i-a_i-t(b_i-a_i))^2=(x_i+(t-1)a_i-tb_i)^2=(x_i-\big{[}(1-t)a_i+tb_i\big{]})$
If I could raise the absolute value in $(...)$ above then I could maybe show with what I already know that the new raised Expression is $<r$. But I don't know how.
Can someone give me a hint how I can solve this excercise please?
If $\lVert a\rVert,\lVert b\rVert<r$, then for each $t\in[0,1]$,\begin{align}\bigl\lVert a+t(b-a)\bigr\rVert&=\bigl\lVert(1-t)a+tb\bigr\rVert\\&\leqslant(1-t)\lVert a\rVert+t\lVert b\rVert\\&<(1-t)r+tr\\&=r.\end{align}If you have $\leqslant $ instead of $<$, it is similar.