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Here is one example. Let $(q_n)_n$ be a list containing all rationals in $[0,1]$, let $f_n:[0,1]\to\mathbb{R}$ be given by $f_n(x)=1$ if $|x-q_n|<\frac{1}{n!}$ and $f_n(x)=0$ if not. Let $f=\sum_nf_n: …

Let $f$ be non linear in the interval $[-1,1]$. We can suppose $f(-1)=f(1)=0$ and $f(x)>0$ for some $x\in(-1,1)$. Now let $N$ be so big that the closed ball of center $(0,-N)$ and with radius $\sqrt{N …

Here is an example of a convex subset $X$ of an infinite-dimensional separable Hilbert space $H$ with empty interior and which is not contained in any hyperplane of $H$, closed or not. Let $(v_i)_{i\i …

To give a positive answer to the question it is enough to, for a fixed $\varepsilon$, give a collection of disjoint balls in $C[0,1]$ of radius $\varepsilon$ which is dense in $C[0,1]$. Indeed, then f …

For any Banach space $X$ you can consider $X\oplus l^2$, with norm $||(x,y)||:=(||x||^2+||y||^2)^\frac{1}{2}$. Then for each $x\in X$, span$(x)\oplus l^2$ is isometric to $l^2$, so $X\oplus l^2$ is co …

A convex $O'$ need not exist: a counterexample is given by setting $K=[-1,1]\times[0,2]\subseteq\mathbb{R}^2$ and $O=\{(x,y)\in K;y>x^3\}$. Indeed, any open $O'$ with $O'\cap K=O$ would contain some n …

If I have understood the definitions correctly, $f(K)$ need not be doubling. For example consider a map $f$ from $[0,1]$ to the Hilbert space $\mathbb{R}\times l^2$ defined in the following way. Let $ …

This is not an answer to the question, but a proof of something much weaker: there is a sequence $c_m\to0$ such that if $\lVert g_m\rVert\leq c_m$ for all $m$, then the result from the question is tru …

The Stone Weierstrass theorem for $C([0,1])$ claims that for any continuous function $f:[0,1]\to\mathbb{R}$ and each $n\in\mathbb{N}$, there is a polynomial $p_{n,f}(x)=\sum_ia_{f,n,i}x^i$ such that $ …

The statement is false in general, I added a counterexample at the end of my answer to show that some separability condition like countable separability (or the stronger condition of being Lebesgue fr …

Seems like in fact convex sets are the only ones for which $p_C$ is continuous. To prove this, we can begin by noticing that for a set $C$ with the property, $p_C$ can only take one point sets as valu …

As you say, it is not difficult to prove that $g=0$ when there are finitely many fluctuations in sign. More precisely, suppose we have points $0=a_0,a_1,\dots,a_n=M$ such that $g\geq0$ or $g\leq0$ in …

Such a function doesn't exist for some choices of $a$. For notational purposes I will change the unit interval by $[-2,2]$. Consider a $C^\infty$ function $a:[-2,2]\to\mathbb{R}$ such that $a(0)=3$, $ …