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For any $n \in \mathbb{N}$, $a \in \mathbb{R}$ with $a > 1$ and $k_i > 0$ for $i = 1,\ldots,n$ define the following function: $$f: \mathbb{R}_{>0}^n \to \mathbb{R}, x \mapsto \sum_{i=1}^n (x_i - k_i)^a$$

I am trying to prove that $f$ is a strictly convex function, i.e. for any $t \in (0,1)$ and for all $x,y\in \mathbb{R}^n$ with $x \neq y$ we have $f(tx+(1-t)y) < tf(x) + (1-t)f(y)$.

Using the definition of the function we have to show that

$$\sum_{i = 1}^n \left( t(x_i-k_i)^a + (1-t)(y_i-k_i)^a - (tx_i + (1-t)y_i -k_i)^a \right) > 0.$$

Since we can set $x_i = y_i = k_i$ for all but one term of the sum, it suffices to look at the single terms and show that they are strictly positive:

$$ t(x_i-k_i)^a + (1-t)(y_i-k_i)^a - (tx_i + (1-t)y_i -k_i)^a > 0.$$

My next idea was to use the Binomial theorem for general exponents, but it didn't lead me anywhere. Does anyone have an idea? It seems to be a well-known statement about distance functions. Thanks!

KDuck
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2 Answers2

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Your function $f$ is actually the $a-$ power of the $a-$ norm in $\mathbb R^n$ of $x-k$: $$f(x)=\|x-k\|_{l^a}^a=\sum\limits_{i=1}^{n}{(x_i-k_i)^a}=(\|x-k\|_{l^a})^a=g(d(x))$$ where $d:\mathbb R^n\to\mathbb R_+$ defined by $$d(x)=\|x-k\|_{l^a}=\left(\sum\limits_{i=1}^{n}{(x_i-k_i)^a}\right)^{1/a}$$ and $g:\mathbb R_+\to\mathbb R_+$ is defined by $g(x)=x^a$ is convex and nondecreasing. We want to show convexity of the composition $f=g\circ d$, i.e $$\forall t\in (0,1):\, f(tx+(1-t)y)\leq tf(x)+(1-t)f(y)$$ But we know that $d$ is convex- just apply triangle inequaliy: $$d(tx+(1-t)y)=\|tx+(1-t)y-k\|_{l^a}=\|tx-tk+(1-t)y-(1-t)k\|_{l^a}$$ $$\leq t\|x-k\|_{l^a}+(1-t)\|y-k\|_{l^a}=td(x)+(1-t)d(y)$$ Now we apply first the monotonicity of $g$ and then its convexity to conclude that $f$ is convex: $$f(tx+(1-t)y)=g(d(tx+(1-t)y))\stackrel{monotonicity}{\leq} g(td(x)+(1-t)d(y))$$ $$\stackrel{convexity}{\leq} tg(d(x))+(1-t)g(d(y))=tf(x)+(1-t)f(y)$$

In conclusion, $f$ is convex for all $x\in\mathbb R^n$ and any vector $k\in\mathbb R^n$ no matter whether $x,k$ are with positive components or not.

Svetoslav
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Just consider $f(x_i) =(x_i - k_i)^\alpha$ for $x_i > k_i$.

$$f'(x_i) = \alpha (x_i - k_i)^{\alpha - 1}$$ $$f''(x_i) = \alpha (\alpha - 1) (x_i - k_i)^{\alpha -2} < 0$$

So f connot be convex for $ 0 < \alpha < 1$, since if f is convex, its derivative should increasing. but $f'' < 0$. f is concave actually.

runaround
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