Show that function $f(x) = 11\|x\|^{22}$ is continuous

Question

Definition for continuity:

Function $f:X\rightarrow Y$ is continuous at $x_0\in X$ if for all $\varepsilon>0$ there exists a $\delta>0$ such that $$ ||x-x_0||<\delta \implies \|f(x)-f(x_0)\| < \varepsilon. $$ Using the given definition, how to show that a function $f:\mathbb{R}^n \rightarrow \mathbb{R}$, $f(x) = 11\|x\|^{22}$ is continuous?

What I have tried so far:

• I noticed that I can write $\|x\|^{22}=\left( \|x\|^{11}\right) ^2$ so I can use identity $(a^2-b^2)=(a+b)(a-b)$.
• I tried to add and subtract same terms.
• Triangle inequality literally in every phase but I never got anything useful there.

I am quite certain that there is a very simple solution for this so any hints are much appreciated!

Answer 1

Apologies for misleading you.

Here is one solution that uses the mean value theorem.

Let $\phi(t) = 11 t^{22}$. Suppose $t_1,t_2 \in [0,T]$, then since $\phi'(t) = 242 \ t^{21}$ we have $|\phi(t_1)-\phi(t_2)| \le 242\ T^{21}|t_1-t_2|$.

Note that $| \|x\| - \|y\| | \le \|x-y\|$.

Pick $x_0$. We can assume that $\delta \le 1$ to start, and so we know $\|x\| \le \|x_0\| + \|x-x_0\| < \|x_0\| +1$.

Now let $\delta = \min(1, {\epsilon \over 242 (\|x_0\|+1)^{21} } ) $.

Then \begin{eqnarray} | 11 \|x\|^{22}-11 \|x_0\|^{22} | &=& |\phi(\|x\|) - \phi(\|x_0\|)| \\ &\le& 242 (\|x_0\|+1)^{21} | \|x\|-\|x_0\|| \\ &\le& 242 (\|x_0\|+1)^{21} \|x-x_0\| \end{eqnarray} and with the chosen $\delta$ we see that if $\|x-x_0\| < \delta$ then $| 11 \|x\|^{22}-11 \|x_0\|^{22} | < \epsilon$.

Answer 2

Here is an answer avoiding $\epsilon-\delta$ techniques. If you insist on an $\epsilon-\delta$ proof you can always look at the $\epsilon-\delta$ proofs of the statements I use in this answer....

Show that $\Vert \cdot \Vert : x \mapsto \Vert x \Vert$ is continuous (this follows immediately by the reverse triangle inequality). Define $g: \mathbb{R} \to \mathbb{R}$ by $g(x) = 11x^{22}, x \in \mathbb{R}$. Then $g$ is continuous, and thus also the composition $g \circ \Vert \cdot \Vert$ is continuous, which is exactly your function.