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Let $f:[0,1] to mathbb R$ be a continuous function and the sequences $(a_n)_n,(b_n)_n$ s.t. $$lim_nto infty int_0^1 |f(x)-a_nx-b_n| dx=0.$$ Prove that $(a_n)_n,(b_n)_n$ are convergent.

I know that $$left|int_0^1f(x)dxright| le int_0^1|f(x)|dx.$$ Can somebody help me, please?

Consider the metric space $E$ of the continuous functions on $[0,1]$, with the distance $d : E times E rightarrow mathbbR_+$ defined by
$$d(f,g)=int_0^1 |f(t)-g(t)| mathrmdt$$

Consider the subspace $A$ of affine functions.

Consider the application $varphi : [0,1]^2 rightarrow E$ defined for all $(a,b) in [0,1]^2$ by
$$varphi(a,b) = lbrace f : t rightarrow (b-a)t+a rbrace$$

Then it is easy to see that $varphi$ is continuous. Moreover, its image is precisely $A$. You deduce that $A$ is a compact subset of $E$ (as a continuous image of the compact $[0,1]^2$), and therefore $A$ is closed.

This proves, in the case of your question, that such a function $f$ has to be affine, as a limit of the closed set of affine functions.

So there exists $a,b$ such that $f(x)=ax+b$. You can rewrite the hypothesis
$$int_0^1 |(a-a_n)x +(b-b_n)| mathrmdx rightarrow 0$$

It is easy now to prove that the only possibility is that $a_n rightarrow a$ and $b_n rightarrow b$.

Here is an elementary proof:

Lemma

If $int_0^1 |f(x)-f_n(x)|dx to 0$ (with $f,f_1,f_2,...$ continuous) and $F(x)=int_0^x f(y)dy,F_n(x)=int_0^x f_n(y)dy$ then $int_0^1 |F(x)-F_n(x)|dx to 0$.

Proof of lemma: $$int_0^1 |F(x)-F_n(x)|dx leq int_0^1int_0^x |f(y)-f_n(y)|dy dx$$ $$ =int_0^1int_y^1 |f(y)-f_n(y)|dxdy =int_0^1 (1-y)|f(y)-f_n(y)|dy leq int_0^1 |f(y)-f_n(y)|dy$$. This proves the lemma.

Now the hypothesis implies $int_0^1 (a_nx+b_n) dx$ converges and the lemma shows that $int_0^1 (a_nx^2/2+b_nx) dx$ also converges. From these two it is quite easy to show that $(a_n)$ and $(b_n)$ both converge.

Another proof using basic measure theory. It is given that $a_nx+b_n to f$ in $L^1$. This implies there is a subsequence which converges almost everywhere, say $a_n_kx+b_n_k to f(x)$ almost everywhere. Form this (using two values of $x$ for which convergence holds) we see that $a_n_k$ and $b_n_k$ both converge, say to $a$ and $b$ so $f(x)=ax+b$ for some $a$ and $b$. Note that this last equation uniquely determines $a$ and $b$. In fact $b=f(0)$ and $a =f(1)-f(0)$ We can now argue that every subsequence of $a_n_k$ has a subsequence converging to the same limit, so $a_n$ is convergent. Similarly $b_n$ is convegent.

We have $|int_0^1 (f(x)-a_nx-b_n) dx| le int_0^1 |f(x)-a_nx-b_n| dx.$

Hence $int_0^1 (f(x)-a_nx-b_n) dx to 0$.

Since $int_0^1 (f(x)-a_nx-b_n) dx = int_0^1 f(x) dx-frac12a_n-b_n$, we see:

$frac12a_n+b_n to int_0^1 f(x) dx$.

Can you proceed ?

Let be $L_n(x) = a_n x + b_n$. The hypothesis says that $L_nto f$ in $L^1$. Consider the vector subspace generated by the polynomials of degree $le 1$ and $f$. Obviously, is finite-dimensional. But in finite dimension all the norms are equivalent, so the convergence $L_nto f$ is uniform...