Cfrgtkky

I am trying to prove that if a function $f : mathbb{R} to mathbb{R}$ can be approximated arbitrarily well by polynomials of bounded degree, then $f$ itself must be a polynomial.

For starters, let ${ g_m mid g_m : mathbb{R} to mathbb{R}}_{m=1}^infty$ be a sequence of polynomials with degree at most $d$ which uniformly converges to $f$. And let the coefficients of the $m$'th approximator be $a_{mn}$ so that $$
g_m(x) = sum_{n=0}^d a_{mn} , x^n.
$$

By "uniform convergence", I mean that given $epsilon > 0$, there exists an $M_epsilon$ such that for all $m ge M_epsilon$ we have $$
sup_{x in mathbb{R}} |f(x) - g_m(x)| < epsilon.
$$

Clearly, proving the claim is equivalent to proving that for each $n$ between 0 and $d$, the sequence of coefficients $a_{1n}, a_{2n}, ldots$ converges to a some limiting value $a_n^*$.

However, it's not clear to me how to prove this statement. Any hints or guidance is greatly appreciated.

For each $n$ you can find some
$$P_n(x)=a_d^nx^d+...+a_1^nx+a_0^n$$
such that
$$sup_{x in mathbb{R}} |f(x) - P_n(x)| < frac{1}{n}$$

Then, for each $n,m$ you have
$$sup_{x in mathbb{R}} |P_m(x) - P_n(x)| leq sup_{x in mathbb{R}} |f(x) - P_n(x)| +sup_{x in mathbb{R}} |f(x) - P_m(x)| < frac{1}{n}+ frac{1}{m}$$

This shows that $P_m(x)-P_n(x)$ is a polynomial which is bounded on $mathbb R$ and hence a constant. Therefore, there exists some polynomial $P(x)$ and some $c_n in mathbb R$ such that
$$P_n(x)=P(x)+c_n$$

By the above you get
$$left| c_n -c_m right| =sup_{x in mathbb{R}} |P_m(x) - P_n(x)| < frac{1}{n}+ frac{1}{m} $$

Therefore, $c_n$ is Cauchy and hence convergent to some $c$.

We claim that $f(x)=P(x)+c$.

Let $epsilon >0$. Pick some $N$ such that, for all $n >N$ we have
$$frac{1}{n} < frac{epsilon}{2}\
|c_n-c|< frac{epsilon}{2}$$

Let $n >N$ be fixed but arbitrary. Then
$$sup_{x in mathbb{R}} |f(x) - P(x)| leq sup_{x in mathbb{R}} |f(x) - P_n(x)| +sup_{x in mathbb{R}} |P_n(x) - P(x)| < frac{1}{n}+ |c_n-c| <frac{epsilon}{2}+frac{epsilon}{2}=epsilon$$

This shows that
$$sup_{x in mathbb{R}} |f(x) - P(x)| < epsilon$$
for all $epsilon >0$.

First pick a polynomial $g(x)$ for $varepsilon = 1$, i.e., $suplimits_{xin mathbb{R}}|f(x)-g(x)|< 1$.
Let $h(x)= f(x)-g(x)$.
Then the same condition holds for $h(x)$:
To obtain an approximation with error $varepsilon$, find one for $f(x)$ and subtract $g(x)$.

So it is enough to verify the assertion for bounded functions.
But that is trivial: a bounded function with this property must be constant.
Indirect proof: if it had two different values (say at points $x,y$), let $delta=|h(x)-h(y)|$, and put $varepsilon:=delta/3$. The approximating polynomial must be constant (otherwise we have a problem with the limit at infinity), but no constant is close enough at both $x$ and $y$.
Hence, $h(x)$ is a constant, and then $f(x)= g(x)+h(x)$ is a polynomial.

Since $f$ is a uniform limit of polynomials (contiunous functions), in particular it is continuous.

Consider the space $C(mathbb{R})$ of continuous functions $mathbb{R} to mathbb{R}$ and equip it with a family of seminorms $|cdot|_{infty, [a,b]}$ given by
$$|g|_{infty, [a,b]} = sup_{xin[a,b]}|g(x)|$$

for all segments $[a,b] subseteq mathbb{R}$.

This turns $C(mathbb{R})$ into a locally convex topological vector space.

Assume that $f$ can be uniformly approximated by polynomials $(p_k)_k$ of degree $le n$.

The subspace $mathbb{R}_{le n}[x]$ of real polynomials of degree $le n$ is a finite dimensional subspace of $C(mathbb{R})$ and hence it is closed in $C(mathbb{R})$. Since $p_k to f$ uniformly on $mathbb{R}$, in particular $|f -p_k|_{infty, [a,b]} to 0$ for all segments $[a,b] subseteq mathbb{R}$.

Hence $p_k to f$ in the above topology so $f$ is in the closure of $mathbb{R}_{le n}[x]$. Since the subspace is closed, we conclude $f in mathbb{R}_{le n}[x]$.

Here is a different approach: Denote $mathcal{B} [-t,t]$ the space of bounded functions on the interval $[-t,t]$. We consider this space with the supremum norm $Vert . Vert _infty $, which makes it a normed space. Denote the space of polynomials with degree at most $d$ by $W=mathbb{R}_{leq d} [x] $, these are continues thus bounded function on the interval $[-t,t]$. But we know more than that: W is a finite dimensional subspace of $mathcal{B} [-t,t]$ for every t, thus it's closed.

If you can approximate $f:mathbb{R} rightarrow mathbb{R}$ aritrarily well on $mathbb{R}$ by elements in $W$, of course you can approximate it arbitrarily well in $mathcal{B} [-t,t]$ for every $t>0$ (and $f in mathcal{B} [-t,t]$). But $W$ is closed and thus $f vert _{[-t,t]} in W$ for all $t$, i.e. it is a polynomial on every $[-t,t]$. But every two polynomials that agree on an open set must be the same polynomial (they have the same derivatives, which give the coefficients, i.e. Taylor Expansion), thus we see $f$ is a polynomial on the whole of $mathbb{R}$