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Let's denote $\Pi_n$ the vector space of polynomial functions on I:=[-1,1] of degree less than or equal to n. The optimization problem you quoted is equivalent to:

Problem(2): Find $q\in\Pi_{n-1}$ that minimizes the uniform distance on I from the function $f(x)=x^n.$

While it is clear by compactness that Problem(2) has a solution, it's not obvious that the solution is unique, because the uniform norm is not uniformly convex (lack of unicity already appears in the analogous problem of point-line distance in $\mathbb{R}^2$ with the max norm). The magic of polynomials is that the solution is always unique, for any continuous function $f$. Not only, but Chebyshev also gave a characterization of the minimizer: it is the unique polynomial $q$ such that $f(x)-q(x)$ attains the maximum absolute value in at least n points, with alternate sign. (Polynomials are not the unique functions with this property; more generally one defines "Haar families", that span finite dimensional spaces analogous to the $\Pi_n$, and for them the argument works as well). Going back to your optimization problem, you have that $p(x)$ is the unique monic polynomial with all real simple zeros and that reaches the maximum absolute value with alternate sign between consecutive zeros. With a bit more work one arrives to $T_n/2^{n-1}$ (or, at this time, one pulls it out of the hat, but at this point I'd consider that quite more fair, and would have no objections, especially if the hat is the most noble one of Pafnuty Lvovich).

PS: Trying to answer the "psychology of mathematics" part of your question (how one arrives to the $T_n$ from the optimization problem). Once the problem is reduced to that of determining the n-th degree polynomial $T$ (say with positive leading coefficient) that oscillates n+1 times between it maximum value 1 and its minimum value -1 on the interval [-1,1], I guess that very soon one suspects that circles and trigonometric functions are around (I can't say it certainly, because I already know the answer!). But if one computes the first few such polynomials, and look at their graphs, they clearly look like sinusoidal curves drawn on a cylinder, and at that point one could recall from high school memories that cos(nx) is a trigonometric polynomial of cos(x), and observe it has clearly the wanted property.

Let's denote $\Pi_n$ the vector space of polynomial functions on $I:=[-1,1]$ of degree less than or equal to $n.$ The optimization problem you quoted is equivalent to:

Problem(2): Find $q\in\Pi_{n-1}$ that minimizes the uniform distance on I from the function $f(x)=x^n.$

While it is clear by compactness that Problem(2) has a solution, it's not obvious that the solution is unique, because the uniform norm is not uniformly convex (lack of unicity already appears in the analogous problem of point-line distance in $\mathbb{R}^2$ with the max norm). The magic of polynomials is that the solution is always unique, for any continuous function $f$. Not only, but Chebyshev also gave a characterization of the minimizer: it is the unique polynomial $q$ such that $f(x)-q(x)$ attains the maximum absolute value in at least n points, with alternate sign. (Polynomials are not the unique functions with this property; more generally one defines "Haar families", that span finite dimensional spaces analogous to the $\Pi_n$, and for them the argument works as well). Going back to your optimization problem, you have that $p(x)$ is the unique monic polynomial with all real simple zeros and that reaches the maximum absolute value with alternate sign between consecutive zeros. With a bit more work one arrives to $T_n/2^{n-1}$ (or, at this time, one pulls it out of the hat, but at this point I'd consider that quite more fair, and would have no objections, especially if the hat is the most noble one of Pafnuty Lvovich).

PS: Trying to answer the "psychology of mathematics" part of your question (how one arrives to the $T_n$ from the optimization problem). Once the problem is reduced to that of determining the $n$-th degree polynomial $T$ (say with positive leading coefficient) that oscillates n+1 times between it maximum value $1$ and its minimum value $-1$ on the interval $[-1,1],$ I guess that very soon one suspects that circles and trigonometric functions are around (I can't say it certainly, because I already know the answer!). But if one computes the first few such polynomials, and look at their graphs, they clearly look like sinusoidal curves drawn on a cylinder, and at that point one could recall from high school memories that $\cos(nx)$ is a trigonometric polynomial of $\cos(x),$ and observe it has clearly the wanted property.

Let's denote $\Pi_n$ the vector space of polynomial functions on I:=[-1,1] of degree less than or equal to n. The optimization problem you quoted is equivalent to:

Problem(2): Find $q\in\Pi_{n-1}$ that minimizes the uniform distance on I from the function $f(x)=x^n.$

While it is clear by compactness that Problem(2) has a solution, it's not obvious that the solution is unique, because the uniform norm is not uniformly convex (lack of unicity already appears in the analogous problem of point-line distance in $\mathbb{R}^2$ with the max norm). The magic of polynomials is that the solution is always unique, for any continuous function $f$. Not only, but Chebyshev also gave a characterization of the minimizer: it is the unique polynomial $q$ such that $f(x)-q(x)$ attains the maximum absolute value in at least n points, with alternate sign. (Polynomials are not the unique functions with this property; more generally one defines "Haar families", that span finite dimensional spaces analogous to the $\Pi_n$, and for them the argument works as well). Going back to your optimization problem, you have that $p(x)$ is the unique monic polynomial with all real simple zeros and that reaches the values 1 and -1 alternately between consecutive zeros. With a bit more work one arrives to $T_n/2^{n-1}$ (or, at this time, one pulls it out of the hat, but at this point I'd consider that quite more fair, and would have no objections, especially if the hat is the most noble one of Pafnuty Lvovich).

Let's denote $\Pi_n$ the vector space of polynomial functions on I:=[-1,1] of degree less than or equal to n. The optimization problem you quoted is equivalent to:

Problem(2): Find $q\in\Pi_{n-1}$ that minimizes the uniform distance on I from the function $f(x)=x^n.$

While it is clear by compactness that Problem(2) has a solution, it's not obvious that the solution is unique, because the uniform norm is not uniformly convex (lack of unicity already appears in the analogous problem of point-line distance in $\mathbb{R}^2$ with the max norm). The magic of polynomials is that the solution is always unique, for any continuous function $f$. Not only, but Chebyshev also gave a characterization of the minimizer: it is the unique polynomial $q$ such that $f(x)-q(x)$ attains the maximum absolute value in at least n points, with alternate sign. (Polynomials are not the unique functions with this property; more generally one defines "Haar families", that span finite dimensional spaces analogous to the $\Pi_n$, and for them the argument works as well). Going back to your optimization problem, you have that $p(x)$ is the unique monic polynomial with all real simple zeros and that reaches the maximum absolute value with alternate sign between consecutive zeros. With a bit more work one arrives to $T_n/2^{n-1}$ (or, at this time, one pulls it out of the hat, but at this point I'd consider that quite more fair, and would have no objections, especially if the hat is the most noble one of Pafnuty Lvovich).

PS: Trying to answer the "psychology of mathematics" part of your question (how one arrives to the $T_n$ from the optimization problem). Once the problem is reduced to that of determining the n-th degree polynomial $T$ (say with positive leading coefficient) that oscillates n+1 times between it maximum value 1 and its minimum value -1 on the interval [-1,1], I guess that very soon one suspects that circles and trigonometric functions are around (I can't say it certainly, because I already know the answer!). But if one computes the first few such polynomials, and look at their graphs, they clearly look like sinusoidal curves drawn on a cylinder, and at that point one could recall from high school memories that cos(nx) is a trigonometric polynomial of cos(x), and observe it has clearly the wanted property.

Let's denote $\Pi_n$ the vector space of polynomial functions on I:=[-1,1] of degree less than or equal to n. The optimization problem you quoted is equivalent to:

Problem(2): Find $q\in\Pi_{n-1}$ that minimizes the uniform distance on I from the function $f(x)=x^n.$

While it is clear by compactness that Problem(2) has a solution, it's not obvious that the solution is unique, because the uniform norm is not uniformly convex (lack of unicity already appears in the analogous problem of point-line distance in $\mathbb{R}^2$ with the max norm). The magic of polynomials is that the solution is always unique, for any continuous function $f$. Not only, but Chebyshev also gave a characterization of the minimizer: it is the unique polynomial $q$ such that $f(x)-q(x)$ attains the maximum absolute value in at least n points, with alternate sign. (Polynomials are not the unique functions with this property; more generally one defines "Haar families", that span finite dimensional spaces analogous to the $\Pi_n$, and for them the argument works as well). Going back to your optimization problem, you have that $p(x)$ is the unique monic polynomial with all real simple zeros and that reaches the values 1 and -1 alternately between consecutive zeros. With a bit more works one arrives to $T_n/2^{n-1}$ (or, at this time, one pulls it out of the hat, but at this point I'd consider that quite more fair, and would have no objections, especially if the hat is the most noble one of Pafnuty Lvovich).

Let's denote $\Pi_n$ the vector space of polynomial functions on I:=[-1,1] of degree less than or equal to n. The optimization problem you quoted is equivalent to:

Problem(2): Find $q\in\Pi_{n-1}$ that minimizes the uniform distance on I from the function $f(x)=x^n.$

While it is clear by compactness that Problem(2) has a solution, it's not obvious that the solution is unique, because the uniform norm is not uniformly convex (lack of unicity already appears in the analogous problem of point-line distance in $\mathbb{R}^2$ with the max norm). The magic of polynomials is that the solution is always unique, for any continuous function $f$. Not only, but Chebyshev also gave a characterization of the minimizer: it is the unique polynomial $q$ such that $f(x)-q(x)$ attains the maximum absolute value in at least n points, with alternate sign. (Polynomials are not the unique functions with this property; more generally one defines "Haar families", that span finite dimensional spaces analogous to the $\Pi_n$, and for them the argument works as well). Going back to your optimization problem, you have that $p(x)$ is the unique monic polynomial with all real simple zeros and that reaches the values 1 and -1 alternately between consecutive zeros. With a bit more work one arrives to $T_n/2^{n-1}$ (or, at this time, one pulls it out of the hat, but at this point I'd consider that quite more fair, and would have no objections, especially if the hat is the most noble one of Pafnuty Lvovich).