If $f$ and $g$ are bounded on $[a,b]$, prove that $\rVert f+g \lVert_u \leq \lVert f \rVert_u + \lVert g \rVert_u$

Question

If $f$ and $g$ are bounded on $[a,b]$, prove that $\lVert f+g \rVert_u \leq \lVert f\rVert_u + \lVert g\rVert_u$.

Where do I begin? I think that I may have to use a contradiction but I'm really confused as to how I should approach questions like this in general. Step by step explanations are greatly appreciated.

Thanks

Answer 1

$$ \begin{align} \|f\|_u+\|g\|_u &=\sup_{x\in[a,b]}|f(x)|+\sup_{y\in[a,b]}|g(y)|\\ &=\sup_{(x,y)\in[a,b]^2}(|f(x)|+|g(y)|) \end{align} $$ whereas $$ \begin{align} \|f+g\|_u &=\sup_{x\in[a,b]}|f(x)+g(x)|\\ &\le\sup_{x\in[a,b]}(|f(x)|+|g(x)|)\\ &=\sup_{\substack{(x,y)\in[a,b]^2\\x=y}}(|f(x)|+|g(y)|) \end{align} $$ Note that the second supremum is taken over a subset of where the first supremum is taken. Therefore, $$ \begin{align} \|f+g\|_u &\le\sup_{\substack{(x,y)\in[a,b]^2\\x=y}}(|f(x)|+|g(y)|)\\ &\le\sup_{(x,y)\in[a,b]^2}(|f(x)|+|g(y)|)\\[6pt] &=\|f\|_u+\|g\|_u \end{align} $$

Answer 2

By triangular inequality,

$$(\forall x\in [a,b]) \;$$ $$ |f (x)+g (x)|\le |f (x)|+|g (x)|$$ $$\le ||f||+||g||=M$$

$$\implies \sup_{x\in[a,b]}\{|f(x)+g(x)|\}\le M $$

Answer 3

Let $||f||_{\infty}=M$, and $||g||_{\infty}=N$ where the $\infty$ subscript denotes the uniform norm (this is the notation that is most commonly seen in textbooks, and will almost surely be the notation you see in future analysis courses.

Now, to prove the statement, recall the triangle inequality $||a+b||\leq ||a||+||b||$. ALL norms satisfy the triangle inequality. For our problem, we just can plug in directly $M,N$.

$||f+g||_{\infty}\leq ||f||_{\infty}+||g||_{\infty}$ which is just $|M+N|\leq |M|+|N|$ where we use the ordinary triangle equality for real numbers in the last step.