Show that $\| \cdot \|^2: X \to \mathbb R$ is convex

Question

Good morning, I'm trying to prove this theorem.

Let $\langle X, \| \cdot \| \rangle$ be a normed vector space. Show that $\| \cdot \|^2: X \to \mathbb R$ is convex.

Could you please verify whether my attempt is fine or contains logical gaps/errors? Any suggestion is greatly appreciated!

My attempt:

Let $x,y \in X$ and $\lambda \in [0,1]$. We have $\| \lambda x + (1-\lambda) y \| \le \| \lambda x \| + \|(1-\lambda) y \| =|\lambda| \| x \| +$ $|1-\lambda| \| y \| = \lambda \| x \| + (1-\lambda) \| y \|$.

Let $\| x \| =a \ge 0$ and $\| y \| =b \ge 0$. Our task is done if we show that $(\lambda a +(1-\lambda)b)^2 \le$ $\lambda a^2 + (1-\lambda) b^2$, which is equivalent to $(a-b)^2 \lambda \le (a+b)^2$. If $a=b$ then it holds for all $\lambda \in \mathbb R$. If $a\neq b$ then it is equivalent to $\lambda \le 1 +4ab/(a-b)^2$. The last inequality trivially holds for all $\lambda \in [0,1]$. This completes the proof.

Answer 1

You are right that it is sufficient to show that $$ (\lambda a +(1-\lambda)b)^2 \le \lambda a^2 + (1-\lambda) b^2 $$ but the remaining calculation looks wrong. The above inequality can be rearranged to $$ 0 \le \lambda a^2 + (1-\lambda) b^2 - \lambda^2 a^2 - 2\lambda(1-\lambda)ab - (1-\lambda)^2 b^2 \\ \iff 0 \le \lambda(1-\lambda)(a-b)^2 $$ and that holds for $0 \le \lambda \le 1$.

Answer 2

I've just found that $\| \cdot \|^n: X \to \mathbb R$ is convex for all $n \in \mathbb N$ and the proof is by induction. It would be great if someone helps me verify it. Thank you so much!

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My attempt:

The statement holds for $n \in \{0,1\}$. Let it hold for some $k$. Then $$\| \lambda x + (1-\lambda) y \|^k \le \lambda \| x \|^k + (1-\lambda) \| y \|^k$$

We need to prove that it holds for $k+1$, i.e. $$\| \lambda x + (1-\lambda) y \|^{k+1} \le \lambda \| x \|^{k+1} + (1-\lambda) \| y \|^{k+1}$$

Out task is done if we show that $$\left (\lambda \| x \|^k + (1-\lambda) \| y \|^k \right) (\lambda \| x \| + (1-\lambda) \| y \|) \le \lambda \| x \|^{k+1} + (1-\lambda) \| y \|^{k+1}$$

Let $\| x \| =a \ge 0$ and $\| y \| =b \ge 0$.

After simplifying the last inequality, we get $$\lambda (1-\lambda) (a-b)(a^k -b^k) \ge 0$$

This inequality trivially holds for all $\lambda \in [0,1]$. This completes the proof.