Convergence of the sequence $f_{1}\left(x\right)=\sqrt{x}$ and  $f_{n+1}\left(x\right)=\sqrt{x+f_{n}\left(x\right)}$

Let $\left\{ f_{n}\right\}$ denote the set of functions on $[0,\infty)$ given by $f_{1}\left(x\right)=\sqrt{x}$ and $f_{n+1}\left(x\right)=\sqrt{x+f_{n}\left(x\right)}$ for $n\ge1$. Prove that this sequence is convergent and find the limit function.

We can easily show that this sequence is is nondecreasing. Originally, I was trying to apply the fact that “every bounded monotonic sequence must converge” but then it hit me this is true for $\mathbb{R}^{n}$. Does this fact still apply on $C[0,\infty)$, the set of continuous functions on $[0,\infty)$. If yes, what bound would we use?

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Apply the Theorem pointwise. That is, fix $x$ and consider the sequence of numbers $(f_n(x))$. The bound will depend on $x$... – David Mitra Jul 28 '12 at 19:26
What bound though? I couldn't figure out one. – Galois Jul 28 '12 at 19:27

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Surely you could guess the limit $g(x)$ of $(f_n(x))_n$, if there is such a limit, for each $x$. Then check that $f_n(x)\leqslant g(x)$ for every $n$... – Did Jul 28 '12 at 19:37

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I think $1+2\sqrt x$ will do: $\sqrt{x+f_n(x)}\le\sqrt{x+1+2\sqrt x}=\sqrt{(1+\sqrt x)^2}$. – David Mitra Jul 28 '12 at 19:43
@David: Honestly, why not use $g(x)$? – Did Jul 28 '12 at 20:06
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4 Answers

Clearly, for each $x \ge 0$ the sequence $\{f_n(x)\}_n$ is increasing.

Let $$g:[0,\infty) \to \mathbb{R},\ g(x)= \left\{ \begin{array}{cc} 0 & \text{ if } x=0\cr \frac{1+\sqrt{1+4x}}{2} & \text{ if } x>0 \end{array} \right..$$

I claim that $f_n(x) \le g(x)$ for every $x \ge 0$. Indeed $f_n(0)=0=g(0)$ for every $n$, and $f_1(x) \le g(x)$ for every $x > 0$. If I suppose that $f_n(x) \le g(x)$ for $n \ge 1$ and $x > 0$, then $$f_{n+1}^2(x)-g^2(x)=x+f_n(x)-\frac{1+2x+\sqrt{1+4x}}{2}=f_n(x)-g(x)\le 0,$$ i.e. $f_{n+1}(x)\le g(x)$. Hence $f_n(x)\le g(x)$ for every $n \ge 1$ and $x > 0$.

Thus for each $x \ge 0$, the sequence $\{f_n(x)\}_n$ is convergent (being increasing and bounded from above), and its (pointwise) limit $f: [0,\infty) \to \mathbb{R}, x \mapsto f(x)$ satisfies $f(x)=\sqrt{x+f(x)}$, i.e. $f(x)=g(x)$ for every $x \ge 0$.

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How did you find $g(x)$? – Galois Jul 28 '12 at 22:55
For $x>0, g(x)$ is the positive solution of $y=\sqrt{x+y}$. – Mercy Jul 29 '12 at 11:56

You know that the sequence is increasing. Now notice that if $f(x)$ is the positive solution of $y^2=x+y$, we have $f_n(x)<f(x) \ \forall n \in \mathbb{N}$. In fact, notice that this hold for $n=1$ and assuming for $n-1$ we have $$f^{2}_{n+1}(x) = x +f_n(x) < x+f(x) = f^{2}(x).$$ Then there exist $\lim_n f_n(x)$ and taking the limit in $f^{2}_{n+1}(x) = x + f_n(x)$, we find that $\lim_n f_n(x)= f(x)$.

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In a sense, the answer is "yes". You can apply the Monotone Convergence Theorem pointwise. That is, for a fixed value of $x$, show that the sequence of nonnegative numbers $\bigl(f_n(x)\bigr)$ is bounded above and increasing. Then of course for each $x$, the sequence $\bigl(f_n(x)\bigr)$ converges to some $f(x)$. (You can find $f(x)$ explicitly by taking the limits of both sides of your defining relation for $f_n$.)

Towards showing that $\bigl(f_n(x)\bigr)$ is indeed bounded above, try using the bound $1+2\sqrt x$.

Perhaps better, as suggested by Did, would be to formally find the limit fuction first, and then show that this serves as an upper bound.

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Hints:

• For every $x\gt0$, the function $u_x:t\mapsto \sqrt{x+t}$ is continuous hence every convergent sequence defined by $x_{n+1}=u_x(x_n)$ for every $n\geqslant0$ has limit $z_x$ such that $u_x(z_x)=z_x$. Here, $z_x=\frac12(1+\sqrt{1+4x})$.
• For every $x\gt0$, $z_x-u_x(t)=c_x(t)\cdot(z_x-t)$, with $c_x(t)=1/(z_x+t)$ hence $0\lt c_x(t)\leqslant 1/z_x$.
• For every $x\gt0$, applying the preceding remark to any sequence $(x_n)_n$ such that $x_0\leqslant z_x$ and $x_{n+1}=u_x(x_n)$ for every $n\geqslant0$ shows that $z_x-z_x^{-n}\cdot(z_x-x_0)\leqslant x_n\leqslant z_x$ for every $n\geqslant0$.
• For every $x\gt0$, $z_x\gt1$.

Conclusion:

• For every $x\gt0$, $f_n(x)\to z_x$. On the other hand, $f_n(0)=0$ for every $n\geqslant0$ hence $f_n(0)\to0\ne z_0$ and the limit is discontinuous at $x=0^+$.
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