My idea is to consider $|x - y| < \delta$, then show that $$|\log(1 + \sqrt{1+x^2}) - \log(1 + \sqrt{1+y^2})| = \bigg|\log\bigg(\dfrac{1 + \sqrt{1+x^2}}{1 + \sqrt{1+y^2}}\bigg)\bigg| < \epsilon$$ But I couldn't find a choice for $x, y$ that could implies the above expression is true. Completing the square doesn't seem to help at all. Any idea? |
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The derivative of $f(x)=\log(1 + \sqrt{1+x^2})$ is $\frac{x}{1 + x^2 + \sqrt{1 + x^2}}$, which is bounded in the whole real line since it is continuous and tends to $0$ as $x\to\pm\infty$. By the Mean Value Theorem, $f$ is Lipschitz and so uniformly continuous. |
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Note that $t\mapsto \log t$ is uniformly continuous on $[1,\infty)$ (proven below). Note also that $t\mapsto 1+\sqrt{1+t^2}$ is uniformly continuous on $\mathbb R$ (also proven below). As the composition of uniformly continuous functions is uniformly continuous, the result follows. To see that $\log$ is uniformly continuous on $[1,\infty)$, fix $\varepsilon>0$. Assume that $1\leq x<y$, $y-x<\delta$. Then $y/x<1+\delta/x\leq1+\delta$, and so $$ \log y-\log x=\log \frac y x\leq\log(1+\delta); $$ if we choose $\delta$ small enough so that $\log(1+\delta)<\varepsilon$, we are done. For the uniform continuity of $g:t\mapsto 1+\sqrt{1+t^2}$, fix $\varepsilon>0$. Choose $x_0$ such that $\sqrt{1+x^2}>3/\varepsilon$ if $|x|\geq x_0$. Since $g$ is continuous on the compact set $[-x_0-1,x_0+1]$, it is uniformly continuous there. So there exists $\delta_1>0$ such that $x,y\in[-x_0,y_0]$ with $|y-x|<\delta_1$ implies $|g(y)-g(x)|<\varepsilon$. Now let $\delta=\min\{\delta_1,\varepsilon/3,1\}$. Suppose that $|x-y|<\delta$. If both $|x|<x_0$, then $|y|<x_0+1$ and so $|g(y)-g(x)|<\varepsilon$ by the uniform continuity on the compact set. If $|x|\geq|x_0$, then $$ |g(y)-g(x)|=|\sqrt{1+y^2}-\sqrt{1+x^2}|\leq|\sqrt{1+y^2}-|y||+||y|-|x||+||x|-\sqrt{1+x^2}|\\ \leq\frac1{|y|+\sqrt{1+y^2}}+|y-x|+\frac1{|x|+\sqrt{1+x^2}}<\frac1{\sqrt{1+y^2}}+\frac1{\sqrt{1+x^2}}+\frac\varepsilon3\\ <\frac\varepsilon3+\frac\varepsilon3+\frac\varepsilon3=\varepsilon. $$ |
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Essentially, what you want to show is that forall $\Delta > 0$, there exists $\delta > 0$ such that $$ \left| \frac{1 + \sqrt{1+x^2}}{1 + \sqrt{1+y^2}} - 1 \right| < \Delta $$ whenever $|x-y| < \delta$. Notice that for $x$ and $y$ with $|x-y| < \delta$, when $x > 0$ is large, $$ 1 \underset{x \to \infty}{\longleftarrow} \frac{ 1 + \sqrt{1+x^2}}{1 + \sqrt{1+(x+\delta)^2}} \le \frac{1 + \sqrt{1+x^2}}{1 + \sqrt{1+y^2}} \le \frac{ 1 + \sqrt{1+x^2}}{1 + \sqrt{1+(x-\delta)^2}} \underset{x \to \infty}{\longrightarrow} 1 $$ and similarly, when $x < 0$ and $-x$ is large, $$ 1 \underset{x \to \infty}{\longleftarrow} \frac{ 1 + \sqrt{1+x^2}}{1 + \sqrt{1+(x-\delta)^2}} \le \frac{1 + \sqrt{1+x^2}}{1 + \sqrt{1+y^2}} \le \frac{ 1 + \sqrt{1+x^2}}{1 + \sqrt{1+(x+\delta)^2}} \underset{x \to \infty}{\longrightarrow} 1 $$ The computations are quite easy ; to compute the limits, just use l'Hospital's rule. Therefore, for every $\Delta > 0$, there exists an $M > 0$ such that $$ \left| \frac{1 + \sqrt{1+x^2}}{1 + \sqrt{1+y^2}} - 1 \right| < \Delta $$ when $x > M$. When $|x-y|< \delta$ with $x \in [-M,M]$, use the continuity of the function $f(x) = 1 + \sqrt{1+x^2}$ on the interval $[-M,M]$ to deduce that it is uniformly continuous, hence that there exists $\delta < 0$ such that $|x-y| < \delta$ implies $$ \left| 1 + \sqrt{1+x^2} - \left( 1 + \sqrt{1 + y^2} \right) \right| = |f(x) - f(y)| < \Delta < \Delta \left( 1 + \sqrt{1 + y^2} \right), $$ hence that $$ \left| \frac{1 + \sqrt{1+x^2}}{1 + \sqrt{1+y^2}} - 1 \right| < \Delta. $$ By continuity of the logarithm function, for every $\varepsilon > 0$, there exists $\Delta > 0$ such that $|\log(z) - \log(1)| < \varepsilon$ whenever $|z-1| < \Delta$. But we have shown that for every $\Delta > 0$ there is a $\delta > 0$ such that our ratio (call it $z(x,y)$) satisfies $|z(x,y) - 1| < \Delta$ whenever $|x-y| < \delta$, hence $|\log(z(x,y)) - 1| < \varepsilon$, which by your calculation gives us uniform continuity. Hope that helps, |
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