In the realm of mathematical analysis, the concept of uniform continuity is crucial. Unlike standard continuity, which is defined at a single point, uniform continuity applies across the entire domain of a function. For a function to be Uniformly continuous, the change in function values must be bounded by the change in input values, regardless of where in the domain these values are chosen. Let’s delve into why the function $f(x) = x^{4}$ fails to meet this criterion, demonstrating it is not uniformly continuous.
A key characteristic we can exploit to understand this lack of uniform continuity comes from algebraic manipulation. Consider the absolute difference between $f(x)$ and $f(y)$:
|x⁴ - y⁴| = |(x² - y²)(x² + y²)| = |(x - y)(x + y)(x² + y²)| = |x - y| ⋅ |x + y| ⋅ |x² + y²|This factorization reveals that even when $x$ and $y$ are arbitrarily close (making $|x – y|$ small), the overall difference $|x⁴ – y⁴|$ can become significantly large if $x$ and $y$ themselves are far from zero. The terms $|x + y|$ and $|x² + y²|$ grow as $x$ and $y$ move away from zero, thus amplifying even a small difference in $|x – y|$. This behavior is a strong indicator of the function not being uniformly continuous.
To formally demonstrate this, we can employ a proof by contradiction, often utilizing the epsilon-delta definition of uniform continuity. Let’s assume, for the sake of contradiction, that $f(x) = x^{4}$ is uniformly continuous. This would imply that for any chosen $varepsilon > 0$, there exists a $delta > 0$ such that for all $x, y$ in the domain, if $|x – y| < delta$, then $|f(x) – f(y)| < varepsilon$.
However, let’s examine a specific case to show this assumption breaks down. Consider $x = frac{delta}{2} + frac{1}{delta}$ and $y = frac{1}{delta}$. Clearly, $|x – y| = |(frac{delta}{2} + frac{1}{delta}) – frac{1}{delta}| = |frac{delta}{2}| = frac{delta}{2} < delta$. If $f(x)$ were uniformly continuous, we should expect $|f(x) – f(y)|$ to be less than some chosen $varepsilon$. Let’s calculate $|f(x) – f(y)|$:
|f(\frac{\delta}{2} + \frac{1}{\delta}) - f(\frac{1}{\delta})| = |(\frac{\delta}{2} + \frac{1}{\delta})^{4} - \frac{1}{\delta^{4}}|Expanding and simplifying this expression:
|(\frac{\delta}{2} + \frac{1}{\delta})^{4} - \frac{1}{\delta^{4}}| &= |\frac{\delta}{2}(\frac{\delta}{2} + \frac{2}{\delta})((\frac{\delta}{2} + \frac{1}{\delta})^{2} + \frac{1}{\delta^{2}})| \\
&= |(\frac{\delta^{2}}{4} + 1)(\frac{\delta^{2}}{4} + 2⋅\frac{\delta}{2}⋅\frac{1}{\delta} + \frac{1}{\delta^{2}} + \frac{1}{\delta^{2}})| \\
&= |(\frac{\delta^{2}}{4} + 1)(\frac{\delta^{2}}{4} + 1 + \frac{2}{\delta^{2}})| \\
&= |\frac{\delta^{4}}{16} + \frac{\delta^{2}}{4} + \frac{1}{2} + \frac{\delta^{2}}{4} + 1 + \frac{2}{\delta^{2}}| \\
&= |\frac{\delta^{4}}{16} + \frac{\delta^{2}}{2} + \frac{2}{\delta^{2}} + \frac{3}{2}|\\
&= \frac{\delta^{4}}{16} + \frac{\delta^{2}}{2} + \frac{2}{\delta^{2}} + \frac{3}{2}As $delta$ approaches 0, the term $frac{2}{delta^{2}}$ becomes arbitrarily large. This means that no matter how small we choose $delta$, we can always find $x$ and $y$ such that $|x – y| < delta$ but $|f(x) – f(y)|$ is greater than any pre-selected $varepsilon$, for instance, we can always make it greater than 3/2.
Therefore, we have reached a contradiction to our initial assumption that $f(x) = x^{4}$ is uniformly continuous. For any $delta > 0$, we can find points $x$ and $y$ within $delta$ of each other for which the function values are not within a fixed $varepsilon$ range. This definitively proves that $f(x) = x^{4}$ is not uniformly continuous on the real numbers. The function’s rate of change is not uniformly bounded, allowing for arbitrarily large changes in function value even for small changes in input, especially as we move away from zero.
