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I would say, monomials are not a Schauder basis for $C[0,1]$ because functions that admits a representation are analytic functions on the unit disc. Actually, $f(t)=\sum_k a_k t^k$ immediately implies differentiability at $t=0,$ which is enough to conclude. So, it's more a matter of non-existence than non-unicity.

With trigonometric polynomials it is a bit more delicate. But the reason is again non-existence of the uniform convergent expansion for some continuous function. Of course, if $f$ admits a representation as uniform limit of a series $\sum_k c_k e^{ikt}$ then the series is its Fourier series, so the claim that $\{ e^{ikt} \}_{k\in\mathbb{Z} }$ is not a Schauder basis for the continuous $2\pi$-periodic functions is equivalent to the statement that a Fourier series of a continuous $2\pi$-periodic function may fail to converge uniformly. (I do not have an example handy; one can also show it indirectly as a consequence of Uniform Boundedness principle if I remember well).

[edit] For instance, Wheeden & Zygmund's book Measure and Integral (p 227) has a nice example of a continuous function $f$ whose Fourier series is unbounded at $0$. The idea is defining $f$ as a uniform limit of a normally convergent series $\sum_{j=1}^{\infty} Q_j(t)$ which is formally made out of an unbounded trigonometric series after a rearrangement and a parenthesization. In other words, each $Q_j$ is a linear combination of $e^{ikt}$ with $k\in I_j,$ and the $I_j$'s are pairwise disjoint finite subsets of $\mathbb{Z}$). So computing the Fourier series of $f,$ one finds again the bad series.

I would say, monomials are not a Schauder basis for $C[0,1]$ because functions that admits a representation are analytic functions on the unit disc. Actually, $f(t)=\sum_k a_k t^k$ immediately implies differentiability at $t=0,$ which is enough to conclude. So, it's more a matter of non-existence than non-uniqueness.

With trigonometric polynomials it is a bit more delicate. But the reason is again non-existence of the uniform convergent expansion for some continuous function. Of course, if $f$ admits a representation as uniform limit of a series $\sum_k c_k e^{ikt}$ then the series is its Fourier series, so the claim that $\{ e^{ikt} \}_{k\in\mathbb{Z} }$ is not a Schauder basis for the continuous $2\pi$-periodic functions is equivalent to the statement that a Fourier series of a continuous $2\pi$-periodic function may fail to converge uniformly. (I do not have an example handy; one can also show it indirectly as a consequence of Uniform Boundedness principle if I remember well).

[edit] For instance, Wheeden & Zygmund's book Measure and Integral (p 227) has a nice example of a continuous function $f$ whose Fourier series is unbounded at $0$. The idea is defining $f$ as a uniform limit of a normally convergent series $\sum_{j=1}^{\infty} Q_j(t)$ which is formally made out of an unbounded trigonometric series after a rearrangement and a parenthesization. In other words, each $Q_j$ is a linear combination of $e^{ikt}$ with $k\in I_j,$ and the $I_j$'s are pairwise disjoint finite subsets of $\mathbb{Z}$). So computing the Fourier series of $f,$ one finds again the bad series.

I would say, monomials are not a Shauder basis for $C[0,1]$ because functions that admits a representation are analytic functions on the unit disc. Actually, $f(t)=\sum_k a_k t^k$ immediately implies differentiability at $t=0,$ which is enough to conclude. So, it's more a matter of non-existence than non-unicity.

With trigonometric polynomials it is a bit more delicate. But the reason is again non-existence of the uniform convergent expansion for some continuous function. Of course, if $f$ admits a representation as uniform limit of a series $\sum_k c_k e^{ikt}$ then the series is its Fourier series, so the claim that $\{ e^{ikt} \}_{k\in\mathbb{Z} }$ is not a Schauder basis for the continuous $2\pi$-periodic functions is equivalent to the statement that a Fourier series of a continuous $2\pi$-periodic function may fail to converge uniformly. (I do not have an example handy; one can also show it indirectly as a consequence of Uniform Boundedness principle if I remember well).

[edit] For instance, Wheeden & Zygmund's book Measure and Integral (p 227) has a nice example of a continuous function $f$ whose Fourier series is unbounded at $0$. The idea is defining $f$ as a uniform limit of a normally convergent series $\sum_{j=1}^{\infty} Q_j(t)$ which is formally made out of an unbounded trigonometric series after a rearrangement and a parenthesization. In other words, each $Q_j$ is a linear combination of $e^{ikt}$ with $k\in I_j,$ and the $I_j$'s are pairwise disjoint finite subsets of $\mathbb{Z}$). So computing the Fourier series of $f,$ one finds again the bad series.

I would say, monomials are not a Schauder basis for $C[0,1]$ because functions that admits a representation are analytic functions on the unit disc. Actually, $f(t)=\sum_k a_k t^k$ immediately implies differentiability at $t=0,$ which is enough to conclude. So, it's more a matter of non-existence than non-unicity.

With trigonometric polynomials it is a bit more delicate. But the reason is again non-existence of the uniform convergent expansion for some continuous function. Of course, if $f$ admits a representation as uniform limit of a series $\sum_k c_k e^{ikt}$ then the series is its Fourier series, so the claim that $\{ e^{ikt} \}_{k\in\mathbb{Z} }$ is not a Schauder basis for the continuous $2\pi$-periodic functions is equivalent to the statement that a Fourier series of a continuous $2\pi$-periodic function may fail to converge uniformly. (I do not have an example handy; one can also show it indirectly as a consequence of Uniform Boundedness principle if I remember well).

[edit] For instance, Wheeden & Zygmund's book Measure and Integral (p 227) has a nice example of a continuous function $f$ whose Fourier series is unbounded at $0$. The idea is defining $f$ as a uniform limit of a normally convergent series $\sum_{j=1}^{\infty} Q_j(t)$ which is formally made out of an unbounded trigonometric series after a rearrangement and a parenthesization. In other words, each $Q_j$ is a linear combination of $e^{ikt}$ with $k\in I_j,$ and the $I_j$'s are pairwise disjoint finite subsets of $\mathbb{Z}$). So computing the Fourier series of $f,$ one finds again the bad series.

I would say, monomials are not a Shauder basis for $C[0,1]$ because functions that admits a representation are analytic functions on the unit disc. Actually, $f(t)=\sum_k a_k t^k$ immediately implies differentiability at $t=0,$ which is enough to conclude. So, it's more a matter of non-existence than non-unicity.

With trigonometric polynomials it is a bit more delicate. But the reason is again non-existence of the uniform convergent expansion for some continuous function. Of course, if $f$ admits a representation as uniform limit of a series $\sum_k c_k e^{ikt}$ then the series is its Fourier series, so the claim that $\{ e^{ikt} \}_{k\in\mathbb{Z} }$ is not a Schauder basis for the continuous $2\pi$-periodic functions is equivalent to the statement that a Fourier series of a continuous $2\pi$-periodic function may fail to converge uniformly. (I do not have an example handy; one can also show it indirectly as a consequence of Uniform Boundedness principle if I remember well).

I would say, monomials are not a Shauder basis for $C[0,1]$ because functions that admits a representation are analytic functions on the unit disc. Actually, $f(t)=\sum_k a_k t^k$ immediately implies differentiability at $t=0,$ which is enough to conclude. So, it's more a matter of non-existence than non-unicity.

With trigonometric polynomials it is a bit more delicate. But the reason is again non-existence of the uniform convergent expansion for some continuous function. Of course, if $f$ admits a representation as uniform limit of a series $\sum_k c_k e^{ikt}$ then the series is its Fourier series, so the claim that $\{ e^{ikt} \}_{k\in\mathbb{Z} }$ is not a Schauder basis for the continuous $2\pi$-periodic functions is equivalent to the statement that a Fourier series of a continuous $2\pi$-periodic function may fail to converge uniformly. (I do not have an example handy; one can also show it indirectly as a consequence of Uniform Boundedness principle if I remember well).

[edit] For instance, Wheeden & Zygmund's book Measure and Integral (p 227) has a nice example of a continuous function $f$ whose Fourier series is unbounded at $0$. The idea is defining $f$ as a uniform limit of a normally convergent series $\sum_{j=1}^{\infty} Q_j(t)$ which is formally made out of an unbounded trigonometric series after a rearrangement and a parenthesization. In other words, each $Q_j$ is a linear combination of $e^{ikt}$ with $k\in I_j,$ and the $I_j$'s are pairwise disjoint finite subsets of $\mathbb{Z}$). So computing the Fourier series of $f,$ one finds again the bad series.