# $\lim\limits_{n\to \infty} \int_0^1 |f(x)-a_nx-b_n| dx=0$ implies $(a_n)_n,(b_n)_n$ are convergent

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

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

Consider the metric space $$E$$ of the continuous functions on $$[0,1]$$, with the distance $$d : E \times E \rightarrow \mathbb{R}_+$$ defined by $$d(f,g)=\int_0^1 |f(t)-g(t)| \mathrm{dt}$$

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 $$\int_0^1 |(a-a_n)x +(b-b_n)| \mathrm{dx} \rightarrow 0$$

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

• Very ingenious argument. +1 – Paramanand Singh Mar 8 at 17:31
• Don't you know a simpler proof?I do not know metric spaces. – Gaboru Mar 8 at 20:29

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: $$\int_0^{1} |F(x)-F_n(x)|dx \leq \int_0^{1}\int_0^{x} |f(y)-f_n(y)|dy dx$$ $$=\int_0^{1}\int_y^{1} |f(y)-f_n(y)|dxdy =\int_0^{1} (1-y)|f(y)-f_n(y)|dy \leq \int_0^{1} |f(y)-f_n(y)|dy$$. 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 $$a_{n_k}x+b_{n_k} \to f(x)$$ 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.

• Why are the sequences $(a_n)$ and $(b_n)$ convergent? – Fred Mar 8 at 21:21
• A sequence $(x_n)$ in a metric space converges to $x$ iff every subsequence of $(x_n)$ has a further subseqeunce which converges to $x$. @Fred – Kavi Rama Murthy Mar 8 at 23:17
• @Gaboru I have added another elementary proof. – Kavi Rama Murthy Mar 8 at 23:37
• Mister, do you know how to proceed Fred's argument? – Gaboru Mar 9 at 17:06
• @Gaboru Fred's answer is incomplete and I have completed it by considering anti-derivatives of $f$ and $a_nx+b_n$. – Kavi Rama Murthy Mar 9 at 23:13

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 $$\int_0^1 (f(x)-a_nx-b_n) dx = \int_0^1 f(x) dx-\frac{1}{2}a_n-b_n$$, we see:

$$\frac{1}{2}a_n+b_n \to \int_0^1 f(x) dx$$.

Can you proceed ?

• But that doesn't imply that both of them are convergent. Because we can be in a infinity-infinity case – Gaboru Mar 8 at 11:58
• I managed to get that alone. I don't know how to proceed. – Gaboru Mar 9 at 17:04

Let be $$L_n(x) = a_n x + b_n$$. The hypothesis says that $$L_n\to f$$ 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 $$L_n\to f$$ is uniform...