If $q:R^n\to R$ is convex and $d^0,\ldots,d^k\in R^n$ are linearly independent, then $R^k\ni l\mapsto q(x^0+\sum_{i=0}^kl_id^i)$ is convex, too

Question

Let $q:\mathbb R^n\to\mathbb R$ be strictly convex, $d^0,\ldots,d^k\in\mathbb R^n$ be linearly independent (for some $k\in\left\{0,\ldots,n-1\right\}$) and $$h:\mathbb R^k\to\mathbb R\;,\;\;\;\lambda\mapsto q\left(x^0+\sum_{i=0}^k\lambda_id^i\right)\;.$$ How can we show, that $h$ is strictly convex?

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On one hand, the statement seems to be trivial, since $$\mathcal Q_{k+1}:=\operatorname{span}\left\{d^0,\ldots,d^k\right\}$$ is a subspace of $\mathbb R^n$ and each $x\in\mathcal Q_{k+1}$ can be uniquely represented in the form $$x=x^0+\sum_{i=0}^k\lambda_id^i$$ for some $\lambda_i\in\mathbb R$. So, the statement should immediately follow, since if $q$ is strictly convex on the whole space $\mathbb R^n$, it's strictly convex on any subspace, too.

On the other hand, it seems that this kind of argumentation is more suitable for the proof that $\tilde{h}:=\left.q\right|_{\mathcal Q_{k+1}}$ is strictly convex.

So, I think I need some argument, that the strict convexity is independent of the choice of the used coordinates.

Answer 1

Hint: Try it from the definition.

$h(p\lambda_1+(1-p)\lambda_2)<ph(\lambda_1)+(1-p)h(\lambda_2)$ Just write $h$ in terms of $q$ in the LHS and apply the definition of convexity. then get back to $h$ from $q$.

Answer 2

Lemma: If $q$ is strictly convex and $\ker A=0$ then $f(x)=q(Ax+b)$ is strictly convex.

Proof: If $x_1\ne x_2$ then $y_1=Ax_1+b\ne y_2=Ax_2+b$ and for $\lambda\in(0,1)$ we have $$ f(\lambda x_1+(1-\lambda)x_2)=q(\lambda y_1+(1-\lambda)y_2)<\lambda q(y_1)+(1-\lambda)q(y_2)=\\=\lambda f(x_1)+(1-\lambda)f(x_2). $$